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PASSAGE - 6
Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of red and black balls, respectively, in box II. (JEE Adv. 2015) Q. One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is 
, then the correct option(s) with the possible values of n1, n2, n3 and n4 is(are) - a)n1 = 3, n2 = 3, n3 = 5, n4 = 15
- b)n1 = 3, n2 = 6, n3 = 10, n4 = 50
- c)n1 = 8, n2 = 6, n3 = 5, n4 = 20
- d)n1 = 6, n2 = 12, n3 = 5, n4 = 20
Correct answer is option 'A,B'. Can you explain this answer?
PASSAGE - 6
Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of red and black balls, respectively, in box II. (JEE Adv. 2015)
Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of red and black balls, respectively, in box II. (JEE Adv. 2015)
Q. One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is , then the correct option(s) with the possible values of n1, n2, n3 and n4 is(are)
, then the correct option(s) with the possible values of n1, n2, n3 and n4 is(are) a)
n1 = 3, n2 = 3, n3 = 5, n4 = 15
b)
n1 = 3, n2 = 6, n3 = 10, n4 = 50
c)
n1 = 8, n2 = 6, n3 = 5, n4 = 20
d)
n1 = 6, n2 = 12, n3 = 5, n4 = 20
|
Krishna Iyer answered |
Let E1 ≡ box I is selected
E2 ≡ box II is selected
E ≡ ball drawn is red
E2 ≡ box II is selected
E ≡ ball drawn is red
P(E2/E) = 
or 
On checking the options we find (a) and (b) are the correct options.
A six faced fair dice is thrown until 1 comes, then the probability that 1 comes in even no. of trials is (2005S)- a)5/11
- b)5/6
- c)6/11
- d)1/6
Correct answer is option 'A'. Can you explain this answer?
A six faced fair dice is thrown until 1 comes, then the probability that 1 comes in even no. of trials is (2005S)
a)
5/11
b)
5/6
c)
6/11
d)
1/6
|
|
Krishna Iyer answered |
In single throw of a dice, probability of getting 1 is = 
and prob. of not getting 1 is 
and prob. of not getting 1 is Then getting 1 in even no. of chances = getting 1 in 2nd chance or in 4th chance or in 6th chance and so on
∴ Req. Prob 
PASSAGE - 3
Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U1 and put into U2.
However, if tail appears then 2 balls are drawn at random from U1 and put into U2 . Now 1 ball is drawn at random from U2. (2011) Q. The probability of the drawn ball from U2 being white is- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
PASSAGE - 3
Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U1 and put into U2.
However, if tail appears then 2 balls are drawn at random from U1 and put into U2 . Now 1 ball is drawn at random from U2. (2011)
Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U1 and put into U2.
However, if tail appears then 2 balls are drawn at random from U1 and put into U2 . Now 1 ball is drawn at random from U2. (2011)
Q. The probability of the drawn ball from U2 being white is
a)
b)
c)
d)
|
Om Rana answered |
PASSAGE - 4
A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls. Q. If 1 ball is drawn from each of the boxes B1, B2 and B3, the probability that all 3 drawn balls are of the same colour is (JEE Adv. 2013)- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
PASSAGE - 4
A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls.
A box B1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3 contains 3 white balls, 4 red balls and 5 black balls.
Q. If 1 ball is drawn from each of the boxes B1, B2 and B3, the probability that all 3 drawn balls are of the same colour is (JEE Adv. 2013)
a)
b)
c)
d)
|
Geetika Shah answered |
Probability that all balls are of same colour = P (all red) + P (all white) + P (all black)
A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is [2007]- a)8/729
- b)8/243
- c)1/729
- d)8/9.
Correct answer is option 'B'. Can you explain this answer?
A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is [2007]
a)
8/729
b)
8/243
c)
1/729
d)
8/9.
|
|
Utkarsh Pandey answered |
Probability of getting score 9 in a single throw
=4/36
=1/9
Probability of getting score 9 exactly twice
=3C2.(1/9)^2.8/9
=8/243
An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5, is then: (1993 - 1 Mark)- a)16 / 81
- b)1 / 81
- c)80 / 81
- d)65 / 81
Correct answer is option 'A'. Can you explain this answer?
An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5, is then: (1993 - 1 Mark)
a)
16 / 81
b)
1 / 81
c)
80 / 81
d)
65 / 81
|
Ankita Das answered |
Option a is correct
Three identical dice are rolled. The probability that the same number will appear on each of them is (1984 - 2 Marks)- a)1/6
- b)1/36
- c)1/18
- d)3/28
Correct answer is option 'B'. Can you explain this answer?
Three identical dice are rolled. The probability that the same number will appear on each of them is (1984 - 2 Marks)
a)
1/6
b)
1/36
c)
1/18
d)
3/28
|
Ananya Das answered |
Favourable cases = 6; {(1, 1, 1), (2, 2, 2),…(6, 6, 6)}
Total cases = 6 × 6 × 6 = 216
∴Req. prob. =
∴Req. prob. =
Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is [2005]- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is [2005]
a)
b)
c)
d)
|
|
Anaya Patel answered |
For a particular house being selected Probability = 
P (all the persons apply for the same house)
Let E and F be two independent events. The probability that exactly one of them occurs is 
and the probability of none of them occurring is 
. If P(T) denotes the probability of occurrence of the event T, then (2011)- a)
- b)
- c)
- d)
Correct answer is option 'A,D'. Can you explain this answer?
Let E and F be two independent events. The probability that exactly one of them occurs is and the probability of none of them occurring is . If P(T) denotes the probability of occurrence of the event T, then (2011)
and the probability of none of them occurring is . If P(T) denotes the probability of occurrence of the event T, then (2011)a)
b)
c)
d)
|
|
Khaja Moinuddin answered |
If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is (1998 - 2 Marks)- a)13/32
- b)1/4
- c)1/32
- d)3/16
Correct answer is option 'A'. Can you explain this answer?
If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is (1998 - 2 Marks)
a)
13/32
b)
1/4
c)
1/32
d)
3/16
|
Ritika Sen answered |
P (2 white and 1 black) = P (W1 W2 B3 or W1 B2 W3 or B1 W2 W3)
= P (W1 W2 B3) + P (W1 B2 W3) + P (B1 W2 W3)
= P (W1) P(W2) P (B3) + P (W1) P(B2) P (W3)
+ P (B1) P (W2) P (W3)
= P (W1 W2 B3) + P (W1 B2 W3) + P (B1 W2 W3)
= P (W1) P(W2) P (B3) + P (W1) P(B2) P (W3)
+ P (B1) P (W2) P (W3)
An experiment has 10 equally likely outcomes. Let A and B be non-empty events of the experiment. If A consists of 4 outcomes, the number of outcomes that B must have so that A and B are independent, is (2008)- a)2, 4 or 8
- b)3, 6 or 9
- c)4 or 8
- d)5 or 10
Correct answer is option 'D'. Can you explain this answer?
An experiment has 10 equally likely outcomes. Let A and B be non-empty events of the experiment. If A consists of 4 outcomes, the number of outcomes that B must have so that A and B are independent, is (2008)
a)
2, 4 or 8
b)
3, 6 or 9
c)
4 or 8
d)
5 or 10
|
Shreya Dasgupta answered |
We have n (S) = 10, n(A) = 4
Let n(B) = x and n(A ∩ B) = y
Then for A and B to be independent events P(A ∩ B) = P(A) P(B)
Let n(B) = x and n(A ∩ B) = y
Then for A and B to be independent events P(A ∩ B) = P(A) P(B)
⇒ y can be 2 or 4 so that x = 5 or 10
∴ n (B) = 5 or 10
∴ n (B) = 5 or 10
The probability of India winning a test match against west Indies is 1/2. Assuming independence from match to match the probability that in a 5 match series India’s second win occurs at third test is (1995S)- a)1/8
- b)1/4
- c)1/2
- d)2/3
Correct answer is option 'B'. Can you explain this answer?
The probability of India winning a test match against west Indies is 1/2. Assuming independence from match to match the probability that in a 5 match series India’s second win occurs at third test is (1995S)
a)
1/8
b)
1/4
c)
1/2
d)
2/3
|
Kirti Datta answered |
Wins at least 3 matches can be calculated using the binomial distribution.
The binomial distribution calculates the probability of getting a certain number of successes (India winning a match) in a fixed number of independent Bernoulli trials (5 matches).
To find the probability that India wins at least 3 matches in a 5-match series, we need to calculate the probabilities of winning 3, 4, and 5 matches and then sum them up.
P(India wins at least 3 matches) = P(India wins 3 matches) + P(India wins 4 matches) + P(India wins 5 matches)
P(India wins k matches) = (5Ck) * (1/2)^k * (1/2)^(5-k)
Where (5Ck) is the binomial coefficient that represents the number of ways to choose k successes from 5 trials.
Calculating each term separately:
P(India wins 3 matches) = (5C3) * (1/2)^3 * (1/2)^(5-3) = 10 * (1/8) * (1/8) = 10/64 = 5/32
P(India wins 4 matches) = (5C4) * (1/2)^4 * (1/2)^(5-4) = 5 * (1/16) * (1/2) = 5/32
P(India wins 5 matches) = (5C5) * (1/2)^5 * (1/2)^(5-5) = 1 * (1/32) * (1/1) = 1/32
Now, summing up the probabilities:
P(India wins at least 3 matches) = 5/32 + 5/32 + 1/32 = 11/32
Therefore, the probability that India wins at least 3 matches in a 5-match series against West Indies is 11/32.
The binomial distribution calculates the probability of getting a certain number of successes (India winning a match) in a fixed number of independent Bernoulli trials (5 matches).
To find the probability that India wins at least 3 matches in a 5-match series, we need to calculate the probabilities of winning 3, 4, and 5 matches and then sum them up.
P(India wins at least 3 matches) = P(India wins 3 matches) + P(India wins 4 matches) + P(India wins 5 matches)
P(India wins k matches) = (5Ck) * (1/2)^k * (1/2)^(5-k)
Where (5Ck) is the binomial coefficient that represents the number of ways to choose k successes from 5 trials.
Calculating each term separately:
P(India wins 3 matches) = (5C3) * (1/2)^3 * (1/2)^(5-3) = 10 * (1/8) * (1/8) = 10/64 = 5/32
P(India wins 4 matches) = (5C4) * (1/2)^4 * (1/2)^(5-4) = 5 * (1/16) * (1/2) = 5/32
P(India wins 5 matches) = (5C5) * (1/2)^5 * (1/2)^(5-5) = 1 * (1/32) * (1/1) = 1/32
Now, summing up the probabilities:
P(India wins at least 3 matches) = 5/32 + 5/32 + 1/32 = 11/32
Therefore, the probability that India wins at least 3 matches in a 5-match series against West Indies is 11/32.
Four fair dice D1, D2, D3 and D4 ; each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously.
The probability that D4 shows a number appearing on one of D1, D2 and D3 is (2012)- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Four fair dice D1, D2, D3 and D4 ; each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously.
The probability that D4 shows a number appearing on one of D1, D2 and D3 is (2012)
The probability that D4 shows a number appearing on one of D1, D2 and D3 is (2012)
a)
b)
c)
d)
|
|
Anaya Patel answered |
D4 can show an umber appearing on one of D1, D2 and D3 in the following cases.
Case I : D4 shows a number which is shown by only one of D1, D2 and D3.
D4 shows a number in 6C1 ways.
One out of D1, D2 and D3 can be selected in 3C1 ways.
The selected die shows the same number as on D4 in one way and rest two dice show the different number in 5 ways each.
∴ Number of ways to happen case I = 6C1 × 3C1 × 1 × 5 × 5 = 450
Case II : D4 shows a number which is shown by only two of D1, D2 and D3.
As discussed in case I, it can happen in the following number of ways = 6C1 × 3C2 × 1 × 1 × 5 = 90
Case III : D4 shows a number which is shown by all three dice D1, D2 and D3.
Number of ways it can be done = 6C1 × 3C3 × 1 × 1 × 1 = 6
∴ Total number of favourable ways = 450 + 90 + 6 = 546
Case I : D4 shows a number which is shown by only one of D1, D2 and D3.
D4 shows a number in 6C1 ways.
One out of D1, D2 and D3 can be selected in 3C1 ways.
The selected die shows the same number as on D4 in one way and rest two dice show the different number in 5 ways each.
∴ Number of ways to happen case I = 6C1 × 3C1 × 1 × 5 × 5 = 450
Case II : D4 shows a number which is shown by only two of D1, D2 and D3.
As discussed in case I, it can happen in the following number of ways = 6C1 × 3C2 × 1 × 1 × 5 = 90
Case III : D4 shows a number which is shown by all three dice D1, D2 and D3.
Number of ways it can be done = 6C1 × 3C3 × 1 × 1 × 1 = 6
∴ Total number of favourable ways = 450 + 90 + 6 = 546
Also total ways = 6 × 6 × 6 × 6
∴ Required Probability = 
and 
then P (B∩C) is (2003S)
Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is [2007]- a)0.2
- b)0.7
- c)0.06
- d)0.14.
Correct answer is option 'D'. Can you explain this answer?
Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is [2007]
a)
0.2
b)
0.7
c)
0.06
d)
0.14.
|
Athul Goyal answered |
Given : Probability of aeroplane I, scoring a target correctly i.e., P(I) = 0.3 probability of scoring a target correctly by aeroplane II, i.e. P(II) = 0.2
∴ The required probability
= 0.7 × 0.2 = 0.14A dice is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is [2002]- a)8/3
- b)3/8
- c)4/5
- d)5/4
Correct answer is option 'D'. Can you explain this answer?
A dice is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is [2002]
a)
8/3
b)
3/8
c)
4/5
d)
5/4
|
Pragati Saha answered |
The event follows binomial distribution with
n = 5, p = 3/6 = 1/2. q = 1 – p = 1/2.;
∴ Variance = npq = 5/4.
n = 5, p = 3/6 = 1/2. q = 1 – p = 1/2.;
∴ Variance = npq = 5/4.
The probability that an event A happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event A happens at least once is (1980)- a)0.936
- b)0.784
- c)0.904
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The probability that an event A happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event A happens at least once is (1980)
a)
0.936
b)
0.784
c)
0.904
d)
none of these
|
|
Rishika Roy answered |
p = 0.4, n = 3, P (X ≥ 1) = ? ⇒ q = 0.6
∴ P (X ≥ 1) = 1– P (X = 0)
= 1– 3C0(0.4)0 (0.6)3 = 1 – 0.216 = 0.784
∴ P (X ≥ 1) = 1– P (X = 0)
= 1– 3C0(0.4)0 (0.6)3 = 1 – 0.216 = 0.784
Let ω be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that ωr1 + ωr2 + ωr3 = 0 is (2010)- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Let ω be a complex cube root of unity with ω ≠ 1. A fair die is thrown three times. If r1, r2 and r3 are the numbers obtained on the die, then the probability that ωr1 + ωr2 + ωr3 = 0 is (2010)
a)
b)
c)
d)
|
Chirag Verma answered |
If ω is a complex cube root of unity then, we know that ω3m +ω3n+1 +ω3p +2 = 0 where m, n, p are integers.
∴ r1, r2, r3 should be of the form 3m, 3n + 1 and 3p + 2 taken in any or d er. As r1 ,r2 ,r3 are the numbers obtained on die, these can take any value from 1 to 6.
∴ m can take values 1 or 2, n can take values 0 or 1p can take values 0 or 1
∴ Number of ways of selecting r1, r2, r3
= 2C1 x 2C1 x 2C1 x 3!.
Also the total number of ways of getting r1, r2, r3 on die = 6 × 6 × 6
∴ r1, r2, r3 should be of the form 3m, 3n + 1 and 3p + 2 taken in any or d er. As r1 ,r2 ,r3 are the numbers obtained on die, these can take any value from 1 to 6.
∴ m can take values 1 or 2, n can take values 0 or 1p can take values 0 or 1
∴ Number of ways of selecting r1, r2, r3
= 2C1 x 2C1 x 2C1 x 3!.
Also the total number of ways of getting r1, r2, r3 on die = 6 × 6 × 6
∴ Required probability=
and 
Which of the following is (are) correct ? (2012)
Two fair dice are tossed. Let x be the event that the first die shows an even number and y be the event that the second die shows an odd number. The two events x and y are :- a)Mutually exclusive (1979)
- b)Independent and mutually exclusive
- c)Dependent
- d)None of these.
Correct answer is option 'D'. Can you explain this answer?
Two fair dice are tossed. Let x be the event that the first die shows an even number and y be the event that the second die shows an odd number. The two events x and y are :
a)
Mutually exclusive (1979)
b)
Independent and mutually exclusive
c)
Dependent
d)
None of these.
|
|
Akshara Deshpande answered |
The two events can happen simultaneously e.g., (2, 3)
∴ not mutually exclusive.
Also are not dependent on each other.
∴ not mutually exclusive.
Also are not dependent on each other.
The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true? (1999 - 3 Marks)- a)p + m + c = 19/20
- b)p + m + c = 27/20
- c)pmc = 1/10
- d)pmc = 1/4
Correct answer is option 'B,C'. Can you explain this answer?
The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c, respectively. Of these subjects, the student has a 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true? (1999 - 3 Marks)
a)
p + m + c = 19/20
b)
p + m + c = 27/20
c)
pmc = 1/10
d)
pmc = 1/4
|
Arka Yadav answered |
According to the problem, m + p + c – mp – mc – pc + mpc = 3/4 …(1)
mp (1– c) + mc (1– p) + pc (1– m) = 2/5 or mp + mc + pc – 3mpc = 2/5 … (2)
Also mp + pc + mc – 2mpc = 1/2 … (3)
mp (1– c) + mc (1– p) + pc (1– m) = 2/5 or mp + mc + pc – 3mpc = 2/5 … (2)
Also mp + pc + mc – 2mpc = 1/2 … (3)
(2) and (3) ⇒ 
∴ mp + mc + pc = 
∴ m + p + c =
Two numbers are selected randomly from the set S = {1, 2, 3, 4, 5, 6} without replacement one by one. The probability that minimum of the two numbers is less than 4 is (2003S)- a)1/15
- b)14/15
- c)1/5
- d)4/5
Correct answer is option 'D'. Can you explain this answer?
Two numbers are selected randomly from the set S = {1, 2, 3, 4, 5, 6} without replacement one by one. The probability that minimum of the two numbers is less than 4 is (2003S)
a)
1/15
b)
14/15
c)
1/5
d)
4/5
|
Akash Das answered |
The minimum of two numbers will be less than 4 if at least one of the numbers is less than 4.
∴ P (at least one no. is < 4), = 1– P (both the no’s are ≥ 4)
Let Ec denote the complement of an event E. Let E, F, G be pairwise independent events with P(G) > 0 and P(E∩F∩G) = 0. Then P(Ec∩ Fc| G) equals (2007 -3 marks)- a)P(Ec) + P(Fc)
- b)P(Ec) – P(Fc)
- c)P(Ec) – P(F)
- d)P(E) – P(Fc)
Correct answer is option 'C'. Can you explain this answer?
Let Ec denote the complement of an event E. Let E, F, G be pairwise independent events with P(G) > 0 and P(E∩F∩G) = 0. Then P(Ec∩ Fc| G) equals (2007 -3 marks)
a)
P(Ec) + P(Fc)
b)
P(Ec) – P(Fc)
c)
P(Ec) – P(F)
d)
P(E) – P(Fc)
|
Siddharth Rane answered |
P( Ec ∩ Fc /G) = P( E ∪ F )c /G) 1 - P(E∪ F /G)
= 1 - P(E / G) - P(F/ G) + P(E∩ F / G)
= 1 - P(E ) -P(F)+O
(∵ E, F, G are pairwise independent and
P(E ∩F ∩G)=0
⇒ P(E ).P(F )=0 as P(G) >0 ) = P(Ec )- P(F)
= 1 - P(E / G) - P(F/ G) + P(E∩ F / G)
= 1 - P(E ) -P(F)+O
(∵ E, F, G are pairwise independent and
P(E ∩F ∩G)=0
⇒ P(E ).P(F )=0 as P(G) >0 ) = P(Ec )- P(F)
One hundred identical coins, each with probability, p, of showing up heads are tossed once. If 0 < p < 1 and the probabilitity of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p is (1988 - 2 Marks)- a)1/2
- b)49/101
- c)50/101
- d)51/101.
Correct answer is option 'D'. Can you explain this answer?
One hundred identical coins, each with probability, p, of showing up heads are tossed once. If 0 < p < 1 and the probabilitity of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p is (1988 - 2 Marks)
a)
1/2
b)
49/101
c)
50/101
d)
51/101.
|
Atharva Chauhan answered |
Prob. of one coin showing head = p
∴ Prob of one coin showing tail = 1– p ATQ coin is tossed 100 times and prob. of 50 coins showing head = prob of 51 coins showing head.
Using binomial prob. distribution
∴ Prob of one coin showing tail = 1– p ATQ coin is tossed 100 times and prob. of 50 coins showing head = prob of 51 coins showing head.
Using binomial prob. distribution
P (X = r)= nCr prqn-r ,
we get, 
⇒ 51 – 51 p = 50 p⇒ 101 p = 51 ⇒ 
If M and N are any two events, the probability that exactly one of them occurs is (1984 - 3 Marks)- a)P(M ) + P( N ) - 2P (M ∩ N)
- b)(P(M ) + P(N) - P (M ∩ N)
- c)P(Mc) + P(Nc) - 2P(Mc ∩ Nc)
- d)P(M ∩ Nc) + P(Mc ∩ N)
Correct answer is option 'A,C,D'. Can you explain this answer?
If M and N are any two events, the probability that exactly one of them occurs is (1984 - 3 Marks)
a)
P(M ) + P( N ) - 2P (M ∩ N)
b)
(P(M ) + P(N) - P (M ∩ N)
c)
P(Mc) + P(Nc) - 2P(Mc ∩ Nc)
d)
P(M ∩ Nc) + P(Mc ∩ N)
|
|
Jatin Iyer answered |
Given that M and N are any two events. To check the probability that exactly one of them occurs.
We check all the options one by one.
(a) P (M) + P (N) – 2 P (M ∩ N) = [P (M) + P (N) – P (M ∩ N)] – P( M ∩ N)
= P (M ∪ N) – P (M ∩ N)
⇒ Prob. that exactly one of M and N occurs.
(b) P (M) + P (N) – P (M ∩ N) = P (M ∪ N)
⇒ Prob. that at least one of M and N occurs.
(c) P (Mc) + P (Nc) – 2 P(Mc ∩ Nc) = 1– P (M) + 1 – P (N) – 2P (M ∪ N)c
= 2 – P (M) – P (N) – 2 [1 – P (M ∪ N)]
= 2 – P(M) – P (N) – 2 + 2 P (M ∪ N)
= P (M ∪ N) + P (M ∪ N) – P(M) – P (N)
= P (M ∪ N) – P (M ∩ N)
⇒ Prob. that exactly one of M and N occurs.
(d) P (M ∩ Nc) + P (Mc ∩ N)
⇒ Prob that M occurs but not N or prob that M does not occur but N occurs.
⇒ Prob. that exactly one of M and N occurs.
Thus we can conclude that (a), (c) and (d) are the correct options.
We check all the options one by one.
(a) P (M) + P (N) – 2 P (M ∩ N) = [P (M) + P (N) – P (M ∩ N)] – P( M ∩ N)
= P (M ∪ N) – P (M ∩ N)
⇒ Prob. that exactly one of M and N occurs.
(b) P (M) + P (N) – P (M ∩ N) = P (M ∪ N)
⇒ Prob. that at least one of M and N occurs.
(c) P (Mc) + P (Nc) – 2 P(Mc ∩ Nc) = 1– P (M) + 1 – P (N) – 2P (M ∪ N)c
= 2 – P (M) – P (N) – 2 [1 – P (M ∪ N)]
= 2 – P(M) – P (N) – 2 + 2 P (M ∪ N)
= P (M ∪ N) + P (M ∪ N) – P(M) – P (N)
= P (M ∪ N) – P (M ∩ N)
⇒ Prob. that exactly one of M and N occurs.
(d) P (M ∩ Nc) + P (Mc ∩ N)
⇒ Prob that M occurs but not N or prob that M does not occur but N occurs.
⇒ Prob. that exactly one of M and N occurs.
Thus we can conclude that (a), (c) and (d) are the correct options.
Let A and B be two events such that 
and 
where 
stands forcomplement of event A. Then events A and B are- a)equally likely and mutually exclusive [2005]
- b)equally likely but not independent
- c)independent but not equally likely
- d)mutually exclusive and independent
Correct answer is option 'C'. Can you explain this answer?
Let A and B be two events such that and where stands forcomplement of event A. Then events A and B are
and where stands forcomplement of event A. Then events A and B area)
equally likely and mutually exclusive [2005]
b)
equally likely but not independent
c)
independent but not equally likely
d)
mutually exclusive and independent
|
|
Hrishikesh Gupta answered |
and 
Also ⇒ P(A ∪ B) = P(A) + P(B) - P(A∩ B)
Hence A and B are independent but not equally likely.
At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10 minute time intervals.
The probability that there is at the most one phone call during a 10-minute time period is [2006]- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10 minute time intervals.
The probability that there is at the most one phone call during a 10-minute time period is [2006]
The probability that there is at the most one phone call during a 10-minute time period is [2006]
a)
b)
c)
d)
|
Sakshi Deshpande answered |
P (at most 1 phone call)
= P(X ≤ 1) = P(X = 0) + P(X= 1)
A random variable X has Poisson distribution with mean 2.
Then P (X > 1.5) equals [2005]- a)
- b)0
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A random variable X has Poisson distribution with mean 2.
Then P (X > 1.5) equals [2005]
Then P (X > 1.5) equals [2005]
a)
b)
0
c)
d)
|
|
Devanshi Bajaj answered |
According to Poission distribution, prob. of getting k successes is
=1- P(x = 0) - P(x=1)
A student appears for tests I, II and III. The student is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in tests I, II and III are p, q and 
respectively. If the probability that the student is successful is , then (1986 - 2 Marks)- a)p = q = 1
- b)p = q =
- c)p = 1, q = 0
- d)p = 1, q =
- e)none of these
Correct answer is option 'C'. Can you explain this answer?
A student appears for tests I, II and III. The student is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in tests I, II and III are p, q and respectively. If the probability that the student is successful is , then (1986 - 2 Marks)
respectively. If the probability that the student is successful is , then (1986 - 2 Marks)a)
p = q = 1
b)
p = q =
c)
p = 1, q = 0
d)
p = 1, q =
e)
none of these
|
|
Sneha Desai answered |
Let A, B, C be the events that the student passes test I, II, III respectively.
Then, ATQ ; P(A) = p; P (B) = q; P(C) =
Then, ATQ ; P(A) = p; P (B) = q; P(C) =
Now the student is successful if A and B happen or A and C happen or A, B and C happen.
ATQ, 
⇒ p + pq = 1 ⇒ p (1 + q) = 1
which holds for p = 1 and q = 0.
Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ? [JEE M 2016]- a)E1 and E3 are independent.
- b)E1, E2 and E3 are independent.
- c)E1 and E2 are independent.
- d)E2 and E3 are independent.
Correct answer is option 'B'. Can you explain this answer?
Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ? [JEE M 2016]
a)
E1 and E3 are independent.
b)
E1, E2 and E3 are independent.
c)
E1 and E2 are independent.
d)
E2 and E3 are independent.
|
Maitri Das answered |
And P(E1 ∩ E2 ∩ E3) = 0 ≠ P (E1) . P(E2) . P(E3)
⇒ E1, E2, E3 are not independent.
is equal to (Here 
are complements of A and B respectively). (1982 - 2 Marks) 
A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant T2 produces 80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that P (computer turns out to be defective given that it is produced in plant T1) = 10P (computer turns out to be defective given that it is produced in plant T2), where P(E) denotes the probability of an event E. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant T2 is (JEE Adv. 2016)- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant T2 produces 80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that P (computer turns out to be defective given that it is produced in plant T1) = 10P (computer turns out to be defective given that it is produced in plant T2), where P(E) denotes the probability of an event E. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant T2 is (JEE Adv. 2016)
a)
b)
c)
d)
|
|
Nidhi Joshi answered |
Also P(D) = 
Let A , B , C be three mutually independent events.Consider the two statements S1 and S2 S1 : A and B ∪ C are independent S2 : A an d B ∩ C are independent Then,(1994)- a)Both S1 an d S2 are true
- b)Only S1 is tr ue
- c)Only S2 is true
- d)Neither S1 nor S2 is true
Correct answer is option 'A'. Can you explain this answer?
Let A , B , C be three mutually independent events.Consider the two statements S1 and S2 S1 : A and B ∪ C are independent S2 : A an d B ∩ C are independent Then,(1994)
a)
Both S1 an d S2 are true
b)
Only S1 is tr ue
c)
Only S2 is true
d)
Neither S1 nor S2 is true
|
Uday Ghoshal answered |
P [A ∩ (B ∪ C)] = P [(A ∩ B) ∪ (A ∩ C)] = P (A ∩ B) + P (A ∩ C) – P (A ∩ B ∩ C)
= P (A) P (B) + P (A) P (C) – P (A) P (B) P(C)
= P (A) [P (B) + P (C) – P (B ∩ C)] = P (A) P (B ∪ C)
= P (A) P (B) + P (A) P (C) – P (A) P (B) P(C)
= P (A) [P (B) + P (C) – P (B ∩ C)] = P (A) P (B ∪ C)
∴ S1 is true.
P (A ∩ (B ∩ C)) = P (A) P(B) P (C) = P (A) P (B ∩ C)
∴ S2 is also true.
P (A ∩ (B ∩ C)) = P (A) P(B) P (C) = P (A) P (B ∩ C)
∴ S2 is also true.
Three numbers are chosen at random without replacement from {1,2,3,..8}. The probability that their minimum is 3, given that their maximum is 6, is : [2012]- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Three numbers are chosen at random without replacement from {1,2,3,..8}. The probability that their minimum is 3, given that their maximum is 6, is : [2012]
a)
b)
c)
d)
|
Arjun Sen answered |
Given sample space = {1,2,3,.....,8}
Let Event A : Maximum of three numbers is 6.
B : Minimum of three numbers is 3.
This is the case of conditional probability
We have to find P (minimum) is 3 when it is given that P (maximum) is 6.
Let Event A : Maximum of three numbers is 6.
B : Minimum of three numbers is 3.
This is the case of conditional probability
We have to find P (minimum) is 3 when it is given that P (maximum) is 6.
Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is [2003]- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is [2003]
a)
b)
c)
d)
|
Nisha Sen answered |
Let 5 horses are H1, H2, H3, H4 and H5. Selected pair of horses will be one of the 10 pairs (i.e.; 5C2 ): H1 H2, H1 H3, H1 H4, H1 H5, H2H3, H2 H4, H2 H5, H3 H4, H3 H5
and H4 H5.
Any horse can win the race in 4 ways.
For example : Horses H2 win the race in 4 ways H1 H2, H2H3, H2H4 and H2H5.
Hence required probability =
and H4 H5.
Any horse can win the race in 4 ways.
For example : Horses H2 win the race in 4 ways H1 H2, H2H3, H2H4 and H2H5.
Hence required probability =
Let A and B be two events such that 
where 
stands for the complement of the event A. Then the events A and B are [JEE M 2014]- a)independent but not equally likely.
- b)independent and equally likely.
- c)mutually exclusive and independent.
- d)equally likely but not independent.
Correct answer is option 'A'. Can you explain this answer?
Let A and B be two events such that where stands for the complement of the event A. Then the events A and B are [JEE M 2014]
where stands for the complement of the event A. Then the events A and B are [JEE M 2014]a)
independent but not equally likely.
b)
independent and equally likely.
c)
mutually exclusive and independent.
d)
equally likely but not independent.
|
Anand Sharma answered |
Given
We know P(A∪ B) = P(A)+ P(B)- P(A∩ B)
∵ P( A) ≠ P(B) so they are not equally likely.
Also 
So A & B are independent.
The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then 
is (1987 - 2 Marks)(Here 
are complements of A and B, respectively).- a)0.4
- b)0.8
- c)1.2
- d)1.4
- e)none
Correct answer is option 'C'. Can you explain this answer?
The probability that at least one of the events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then is (1987 - 2 Marks)
is (1987 - 2 Marks)(Here are complements of A and B, respectively).
are complements of A and B, respectively).a)
0.4
b)
0.8
c)
1.2
d)
1.4
e)
none
|
|
Mihir Sharma answered |
Given that P (A ∪ B) = 0.6 ; P (A ∩ B) = 0.2
= 2 – (P (A) + P (B)) = 2 – [P (A ∪ B) + P (A ∩ B)].
= 2 – [0.6 + 0.2] = 2 – 0.8 = 1.2
= 2 – [0.6 + 0.2] = 2 – 0.8 = 1.2
If 
are the complementary events of events E and F respectively and if 0 < P(F) < 1, then (1998 - 2 Marks)- a)
- b)
- c)
- d)
Correct answer is option 'A,D'. Can you explain this answer?
If are the complementary events of events E and F respectively and if 0 < P(F) < 1, then (1998 - 2 Marks)
are the complementary events of events E and F respectively and if 0 < P(F) < 1, then (1998 - 2 Marks)a)
b)
c)
d)
|
|
Siddharth Basak answered |
We have,
(a) 
∴ (a) holds.
Also
Also
(b) 
∴ (b) does not hold. Similarly we can show that (c) does not hold but (d) holds.
PASSAGE - 5
Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be number on the card drawn from the ith box, i = 1, 2, 3. (JEE Adv. 2014) Q. The probability that x1 + x2 + x3 is odd, is- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
PASSAGE - 5
Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be number on the card drawn from the ith box, i = 1, 2, 3. (JEE Adv. 2014)
Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xi be number on the card drawn from the ith box, i = 1, 2, 3. (JEE Adv. 2014)
Q. The probability that x1 + x2 + x3 is odd, is
a)
b)
c)
d)
|
|
Sparsh Roy answered |
x1 + x2 + x3 will be odd If two are even and one is odd or all three are odd.
∴ Required probability = P (EEO) + P(EOE) + P(OEE) + P(OQO)
A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct.
The probability that a student will get 4 or more correct answers just by guessing is: [JEE M 2013]- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct.
The probability that a student will get 4 or more correct answers just by guessing is: [JEE M 2013]
The probability that a student will get 4 or more correct answers just by guessing is: [JEE M 2013]
a)
b)
c)
d)
|
Aarav Roy answered |
p = P (correct answer), q = P (wrong answer)
By using Binomial distribution
Required probability = 
A random variable X has the probability distribution: 
For the events E = {X is a prime number } and F = {X< 4}, the P( E ∪ F) is [2004]- a)0.50
- b)0.77
- c)0.35
- d)0.87
Correct answer is option 'B'. Can you explain this answer?
A random variable X has the probability distribution: For the events E = {X is a prime number } and F = {X< 4}, the P( E ∪ F) is [2004]
For the events E = {X is a prime number } and F = {X< 4}, the P( E ∪ F) is [2004]a)
0.50
b)
0.77
c)
0.35
d)
0.87
|
Partho Sengupta answered |
P(E) = P ( 2 or 3 or 5 or 7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62
P( F ) =P(1 or2 or3) = 0.15 + 0.23 + 0.12 = 0.50 P( E ∩ F )=P(2 or 3) = 0.23 + 0.12 = 0.35
∴ P(EUF ) = P(E) + P(F ) - P(E∩ F )
= 0.62 + 0.50 - 0.35 = 0.77
P( F ) =P(1 or2 or3) = 0.15 + 0.23 + 0.12 = 0.50 P( E ∩ F )=P(2 or 3) = 0.23 + 0.12 = 0.35
∴ P(EUF ) = P(E) + P(F ) - P(E∩ F )
= 0.62 + 0.50 - 0.35 = 0.77
For any two events A and B in a sample space (1991 - 2 Marks)- a)
is always true - b)
does not hold - c)
, if A and B are independent - d)
, if A and B are disjoint.
Correct answer is option 'A,C'. Can you explain this answer?
For any two events A and B in a sample space (1991 - 2 Marks)
a)
is always trueb)
does not holdc)
, if A and B are independentd)
, if A and B are disjoint.|
|
Isha Choudhury answered |
For any two events A and B
(a)
Now we know P (A ∪ B) ≤ 1 P (A) + P (B) – P (A ∩ B) ≤ 1
⇒ P (A ∩ B) > P (A) + P (B) – 1
⇒ P (A ∩ B) > P (A) + P (B) – 1
∴ (a) is correct statement.(b)
From venn diagram we can clearly conclude that
∴ (b) is incorrect statement.
(c) P (A ∪ B) =P (A) + P (B) – P (A ∩ B)
[ ∵ A & B are independent events]
∴ (c) is the correct statement.(d) For disjoint events P (A ∪ B ) = P (A) + P (B)
∴ (d) is the incorrect statement.
In a binomial distribution 
if the probability of at least one success is greater than or equal to 
then n is greater than: [2009]- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
In a binomial distribution if the probability of at least one success is greater than or equal to then n is greater than: [2009]
if the probability of at least one success is greater than or equal to then n is greater than: [2009]a)
b)
c)
d)
|
|
Simran Chopra answered |
We have
Taking log to the base 3/4, on both sides, we get
A fair coin is tossed repeatedly. If the tail appears on first four tosses, then the probability of the head appearing on the fifth toss equals (1998 - 2 Marks)- a)1/2
- b)1/32
- c)31/32
- d)1/5
Correct answer is option 'A'. Can you explain this answer?
A fair coin is tossed repeatedly. If the tail appears on first four tosses, then the probability of the head appearing on the fifth toss equals (1998 - 2 Marks)
a)
1/2
b)
1/32
c)
31/32
d)
1/5
|
|
Nandita Sharma answered |
Th e even t that th e fifth toss r esults in a h ead is independent of the event that the first four tosses result in tails.
∴ Probability of the required event = 1/2.
∴ Probability of the required event = 1/2.
Two events A an d B ha ve pr obabilities 0.25 an d 0.50 respectively. The probability that both A and B occur simultaneously is 0.14. Then the probability that neither A nor B occurs is (1980)- a)0.39
- b)0.25
- c)0.11
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Two events A an d B ha ve pr obabilities 0.25 an d 0.50 respectively. The probability that both A and B occur simultaneously is 0.14. Then the probability that neither A nor B occurs is (1980)
a)
0.39
b)
0.25
c)
0.11
d)
none of these
|
Lakshya Agnihotri answered |
Yeah its 0.39
solving with van diagram we get as...
Total prob=1
p(a)=.25
p(b)=.50
and p(a intersection b)=.14
thus by van diagram...
1-(0.25+0.50-0.14) bcoz 0.14 was included 2 times... once in 0.25 and the other time in0.50.
ans is 0.39
solving with van diagram we get as...
Total prob=1
p(a)=.25
p(b)=.50
and p(a intersection b)=.14
thus by van diagram...
1-(0.25+0.50-0.14) bcoz 0.14 was included 2 times... once in 0.25 and the other time in0.50.
ans is 0.39
In dia plays two matches each with West In dies and Australia. In any match the probabilities of India getting, points 0, 1 and 2 are 0.45, 0.05 and 0.50 respectively.Assuming that the outcomes are independent, the probability of India getting at least 7 points is (1992 - 2 Marks)- a)0.8750
- b)0.0875
- c)0.0625
- d)0.0250
Correct answer is option 'B'. Can you explain this answer?
In dia plays two matches each with West In dies and Australia. In any match the probabilities of India getting, points 0, 1 and 2 are 0.45, 0.05 and 0.50 respectively.Assuming that the outcomes are independent, the probability of India getting at least 7 points is (1992 - 2 Marks)
a)
0.8750
b)
0.0875
c)
0.0625
d)
0.0250
|
|
Pallabi Basak answered |
P (at least 7 pts) = P (7pts) + P (8 pts) [∵ At most 8 pts can be scored.]
Now 7 pts can be scored by scoring 2 pts in 3 matches and 1 pt. in one match, similarly 8 pts can be scored by scoring 2 pts in each of the 4 matches.
∴ Req. prob. = 4C1 × [P (2 pts)]3 P (1pt) + [P (2 pts)]4 = 4 (0.5)3 × 0.05 + (0.50)4 = 0.0250 + 0.0625 = 0.0875
Now 7 pts can be scored by scoring 2 pts in 3 matches and 1 pt. in one match, similarly 8 pts can be scored by scoring 2 pts in each of the 4 matches.
∴ Req. prob. = 4C1 × [P (2 pts)]3 P (1pt) + [P (2 pts)]4 = 4 (0.5)3 × 0.05 + (0.50)4 = 0.0250 + 0.0625 = 0.0875
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