Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

Maths 35 Years JEE Mains & Advance Past year Papers Class 11

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Q.1. For a biased die the probabilities for the different faces to turn up are given below :

Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

This die is tossed and you are told that either face 1 or face 2 has turned up. Then the probability that it is face 1 is ................... (1981 - 2 Marks)

Ans. Sol. Let E1 ≡ face 1 has turned up,E2 ≡ face 1 or 2 has turned up.
By the given data P (E2) = 0.1 + 0.32 = 0.42, P (E1 ∩ E2) = P(E1) = 0.1
Given that E2 has happened and we have to find the probability of happening of E1.
∴ By conditional probability theorem, we have

Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

 

 

Q.2. P( A ∪ B) = P( A∩B) if and only if th e relation between P(A) and P(B) is ................... (1985 - 2 Marks) 

Ans. Sol. Given that P (A ∪ B) = P (A ∩ B) ⇒ P (A) + P (B) – P (A ∩ B ) = P (A ∩ B)
⇒ [P (A) – P (A ∩ B)] + [P(B) – P (A ∩ B)] = 0
But P (A) – P (A ∩ B), P (B) – P (A ∩ B) ≥ 0 [Q P (A ∩ B) ≤ P (A), P (B)]
⇒ P (A) – P (A ∩ B) = 0 and P (B) – P (A ∩ B) = 0 [Q Sum of two non-negative no’s can be zero only when these no’s are zeros]
⇒ P (A) = P (B) = P (A ∩ B)

 

Q.3. A box contains 100 tickets numbered 1, 2, ... .., 100. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The minimum number on them is 5 with probability ................... (1985 - 2 Marks)

Ans. Sol. Let A be the event that max. number on the two chosen tickets in not more than 10, and B is the event that min. number on them is 5. We have to find P (B / A).
We know that Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

Total ways to select two tickets out of 100 = 100C2.
Number of ways favourable to A = number of ways of selecting any 2 numbers from 1 to 10 = 10C2 = 45
A ∩ B contains one number 5 and other greater than 5 and ≤ 10
So ways favourable to A ∩ B = 5C= 5 Therefore,  Fill in the Blanks: Probability | JEE Advanced Notes | EduRevand  Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

ThusFill in the Blanks: Probability | JEE Advanced Notes | EduRevFill in the Blanks: Probability | JEE Advanced Notes | EduRev

 

 

Q.4. If Fill in the Blanks: Probability | JEE Advanced Notes | EduRevFill in the Blanks: Probability | JEE Advanced Notes | EduRevare the probabilities of three mutually exclusive events, then the set of all values of p is ................... (1986 - 2 Marks) 

Ans. Sol. Let   Fill in the Blanks: Probability | JEE Advanced Notes | EduRev, Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

As A, B and C are three mutually exclusive events

∴ P (A) + P (B) + P (C) ≤ 1

Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

⇒ 4 + 12p + 3 – 3p + 6 – 12p ≤ 12

⇒ 3p ≥ 1 ⇒ p ≥ 1 /3          … (i)

Also  Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

⇒ 0 ≤ 1 + 3p ≤ 3

Fill in the Blanks: Probability | JEE Advanced Notes | EduRev             … (ii)

Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

⇒ 0 ≤ 1 – p ≤ 4

⇒ – 3 ≤ p ≤ 1 … (iii)

0 ≤ P (C) ≤ 1 ⇒ Fill in the Blanks: Probability | JEE Advanced Notes | EduRevFill in the Blanks: Probability | JEE Advanced Notes | EduRev… (iv)

Combining (i), (ii), (iii) and (iv), we get Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

 

 

Q.5. Urn A contains 6 red and 4 black balls and urn B contains 4 red and 6 black balls. One ball is   drawn at random from urn A and placed in urn B. Then one ball is drawn at random from urn B and placed in urn A. If one ball is now drawn at random from urn A, the probability that it is found to be red is ................... (1988 - 2 Marks) 

Ans. Sol. There may be following cases: Case I : Red from A to B and red from B to A then prob. of

drawing a red ball from  Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

Case II : Red from A to B and black from B to A then prob. of drawing a red from Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

Case III : Black from A to B and red from B to A then prob. of drawing red from Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

Case IV : Black from A to B and black from B to A then prob. of drawing red from   Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

∴ The required prob  Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

 

 

Q.6. A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. Then the probability that 5 comes before 7 is ................... (1989 - 2 Marks) 

Ans. Sol. Probability of getting a sum of  Fill in the Blanks: Probability | JEE Advanced Notes | EduRev = P(A) as favourable cases are {(1, 4), (4, 1), (2, 3), (3, 2) } Similarly favourable cases of getting a sum of 7 are {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)} = 6

∴ Prob. of getting a sum of  Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

∴ Prob. of getting a sum of 5 or 7

Fill in the Blanks: Probability | JEE Advanced Notes | EduRev [as events are mutually exclusive.]

∴ Prob of getting neither a sum of 5 nor of 7 Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

Now we get a sum of 5 before a sum of 7 if either we get a sum of 5 in first chance or we get neither a sum of 5 nor of 7 in first chance and a sum of 5 in second chance and so on.
Therefore the required prob. is

Fill in the Blanks: Probability | JEE Advanced Notes | EduRevFill in the Blanks: Probability | JEE Advanced Notes | EduRev

 

 

Q.7. Let A and B be two events such that P (A) = 0.3 and P ( A ∪ B) = 0.8. If A and B are independent events then P (B) = ................... (1990 -  2 Marks) 

Ans. Sol. P (A ∪ B) = 0.8
⇒ P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ P (A ∪ B) = P (A) + P (B) – P (A) P (B)
[As A and B are independent events]
⇒ 0.8 = 0.3 + P (B) – 0.3 P (B)
⇒ 0.5 = 0.7 P (B) ⇒ P (B) = 5/7

 

Q.8. If the mean and the variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than one is equal to ................... (1991 -  2 Marks) 

Ans. Sol. For a binomial distribution, we know, mean = np and variance = npq
∴ np = 2; npq = 1 ⇒ q = 1/2 ⇒ p = 1/2 and n = 4
∴ P (X > 1) = P (X = 2) + P (X = 3) + P (X = 4)
= 1– P (X = 0) – P (X = 1)

Fill in the Blanks: Probability | JEE Advanced Notes | EduRevFill in the Blanks: Probability | JEE Advanced Notes | EduRev

 

 

Q.9. Three faces of a fair die are yellow, two faces red and one blue. The die is tossed three times. The probability that the colours, yellow, red and blue, appear in the first, second and the third tosses respectively is .....................(1992 -  2 Marks) 

Ans. Sol. Sample space = {Y, Y,Y, R, R, B} where Y stands for yellow colour, R for red and B for blue.
Prob. that the colours yellow, red and blue appear in the first second, and third tosses respectively   Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

 

 

Q.10. If two events A and B are such that P(Ac) = 0.3, P(B) = 0.4 and P ( A ∩B c) = 0.5, then P(B/(A ∪ Bc)] = ................... (1994 -  2 Marks)

Ans. Sol. Given that P (Ac) = 0.3, P (B) = 0.4 and P (A ∩ Bc) = 0.5 then P [B/ (A ∩ Bc)] =  Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

Fill in the Blanks: Probability | JEE Advanced Notes | EduRev

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