JEE Advanced (Single Correct Type): Probability | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. What is the total number of sample spaces when a die is thrown 2 times?
(a) 6
(b) 12
(c) 18
(d) 36

Correct Answer is option (d)
The possible outcomes when a die is thrown are 1, 2, 3, 4, 5, and 6.
Given, a die is thrown two times.
Then, the total number of sample spaces = (6 × 6)
= 36

Q.2. P(A ∩ B) is equal to:
(a) P(A) . P(B|A)
(b) P(B) . P(A|B)
(c) Both A and B
(d) None of these

Correct Answer is option (c)
By multiplication theorem of probability. P(A ∩ B) is equal to P(A) . P(B|A) and P(B) . P(A|B).

Q.3. What is the total number of sample spaces when a coin is tossed and a die is thrown
(a) 6
(b) 12
(c) 8
(d) 16

Correct Answer is option (b)
The possible outcomes when a coin is tossed are Head (H) or Tail (T).
The possible outcomes when a die is thrown are 1, 2, 3, 4, 5, and 6.
Then, total number of space = (2 × 6) = 12

Q.4. If P (A) = 0.8, P (B) = 0.5 and P (B|A) = 0.4, what is the value of P (A ∩ B)?
(a) 0.32
(b) 0.25
(c) 0.1
(d) 0.5

Correct Answer is option (a)
Given, P (A) = 0.8, P (B) = 0.5 and P (B|A) = 0.4
By conditional probability, we have;
P (B|A) = P(A ∩ B)/P(A)
⇒ P (A ∩ B) = P(B|A). P(A) = 0.4 x 0.8 = 0.32

Q.5. Three identical dice are rolled. What is the probability that the same number will appear on each of them?
(a) 1/6
(b) 1/36
(c) 1/18
(d) 3/28

Correct Answer is option (b)
Total number of cases = 6³ = 216
The same number can appear on each of the dice in the following ways:
(1, 1, 1), (2, 2, 2), ………….(3, 3, 3)
So, favourable number of cases = 6
Hence, required probability = 6/216 = 1/36

Q.6. If P (A) = 6/11, P (B) = 5/11 and P (A ∪ B) = 7/11, what is the value of P(B|A)?
(a) ⅓
(b) ⅔
(c) 1
(d) None of the above

Correct Answer is option (b)
By definition of conditional probability we know;
P(B|A) = P(A ∩ B)/P(A) …(i)
Also,
P(A ∩ B) = P(A) + P(B) – P(A U B)
= 6/11 + 5/11 – 7/11
= 4/11
Now putting the value of P(A ∩ B) in eq.(i), we get;
P(B|A) = (4/11)/(6/11) = 4/6 = ⅔

Q.7. A bag contains 5 brown and 4 white socks. Ram pulls out two socks. What is the probability that both the socks are of the same colour?
(a) 9/20
(b) 2/9
(c) 3/20
(d) 4/9

Correct Answer is option (d)
Total number of socks = 5 + 4 = 9
Two socks are pulled.
Now, P(Both are same colour) = (5C2 + 4C2)/9C2
= {(5×4)/(2×1) + (4×3)/(2×1)}/{(9×8)/(2×1)}
= {(5×4) + (4×3)/}/{(9×8)
= (5 + 3)/(9×2)
= 8/18
= 4/9

Q.8. Find P(E|F), where E: no tail appears, F: no head appears, when two coins are tossed in the air.
(a) 0
(b) ½
(c) 1
(d) None of the above

Correct Answer is option (a)
Given,
E: no tail appears
And F: no head appears
⇒ E = {HH} and F = {TT}
⇒ E ∩ F = ϕ
As we know, two coins were tossed;
P(E) = ¼
P(F) = ¼
P(E ∩ F) = 0/4 = 0
Thus, by conditional probability, we know that;
P(E|F) = P(E ∩ F)/P(F)
= 0/(¼)
= 0

Q.9. What is the probability of getting the number 6 at least once in a regular die if it can roll it 6 times?
(a) 1 – (5/6)⁶
(b) 1 – (1/6)⁶
(c) (5/6)⁶
(d) (1/6)⁶

Correct Answer is option (a)
Let A be the event that 6 does not occur at all.
Now, the probability of at least one 6 occurs = 1 – PA.
= 1 – (5/6)⁶

Q.10. If P(A ∩ B) = 70% and P(B) = 85%, then P(A/B) is equal to:
(a) 17/14
(b) 14/17
(c) ⅞
(d) ⅛

Correct Answer is option (b)
By conditional probability, we know;
P(A|B) = P(A ∩ B)/P(B)
= (70/100) x (100/85)
= 14/17

Q.11. Events A and B are said to be mutually exclusive if:
(a) P (A U B) = P A. + P B.
(b) P (A ∩ B) = P A. × P B.
(c) P(A U B) = 0
(d) None of these

Correct Answer is option (a)
If A and B are mutually exclusive events,
Then P(A ∩ B) = 0
Now, by the addition theorem,
P(A U B) = PA. + PB. – P(A ∩ B)
⇒ P(A U B) = PA. + PB.

Q.12. If P(A) = 0.4, P(B) = 0.7 and P(B/A) = 0.6. Find P(A ∪ B).
(a) 0.46
(b) 0.86
(c) 0.76
(d) 0.54

Correct Answer is option (b)
Given,
P(A) = 0.4, P(B) = 0.7 and P(B/A) = 0.6
By conditional probability, we know;
P(B|A) = P(A ∩ B)/P(A)
⇒ 0.6 × 0.4 = P (A ∩ B)
⇒ P(A ∩ B) = 0.24
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.4 + 0.7 – 0.24
= 0.86

Q.13. A die is rolled. What is the probability that an even number is obtained?
(a) 1/2
(b) 2/3
(c) 1/4
(d) ¾

Correct Answer is option (a)
When a die is rolled, total number of outcomes = 6 (1, 2, 3, 4, 5, 6)
Total even number = 3 (2, 4, 6)
So, the probability that an even number is obtained = 3/6 = ½

Q.14. An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?
(a) 3/7
(b) 7/3
(c) 1/7
(d) ⅓

Correct Answer is option (a)
Let E and F denote the events that the first and second ball drawn is black, respectively.
We need to find P(E ∩ F) or P (EF).
P(E) is the probability of black ball first drawn.
P(E) = 10/15
Now, 9 black balls are left in the urn.
P(F|E) = 9/14
By multiplication rule;
P (E ∩ F) = P (E) P(F|E)
= 10/15 x 9/14
= 3/7

Q.15. What is the probability of selecting a vowel in the word “PROBABILITY”?
(a) 2/11
(b) 3/11
(c) 4/11
(d) 5/11

Correct Answer is option (b)
PROBABILITY = 3/11

Q.16. If E and F are independent events, then;
(a) P(E ∩ F) = P(E)/ P(F)
(b) P(E ∩ F) = P(E) + P(F)
(c) P(E ∩ F) = P(E) . P(F)
(d) None of the above

Correct Answer is option (c)
Two events E and F are said to be independent if;
P(F|E) = P (F) given P (E) ≠ 0
and
P (E|F) = P (E) given P (F) ≠ 0
Now, by the multiplication rule of probability, we have
P(E ∩ F) = P(E) . P (F|E) … (1)
If E and F are independent, then;
P(E ∩ F) = P(E) . P(F)

Q.17. An urn contains 6 balls of which two are red and four are black. Two balls are drawn at random. What is the probability that they are of different colours?
(a) ⅖
(b) 1/15
(c) 8/15
(d) 4/15

Correct Answer is option (c)
Given that, the total number of balls = 6 balls
Let A and B be the red and black balls, respectively,
The probability that two balls are drawn are different = P (the first ball drawn is red)(the second ball drawn is black)+ P (the first ball drawn is black)P(the second ball drawn is red)
= (2/6)(4/5) + (4/6)(2/5)
=(8/30)+ (8/30)
= 16/30
= 8/15

Q.18. If A and B are two independent events, then the probability of occurrence of at least one of A and B is given by:
(a) 1+ P(A′) P (B′)
(b) 1− P(A′) P (B′)
(c) 1− P(A′) + P (B′)
(d) 1− P(A′) – P (B′)

Correct Answer is option (b)
P(at least one of A and B) = P(A ∪ B)
= P(A) + P(B) − P(A ∩ B)
= P(A) + P(B) − P(A) P(B)
= P(A) + P(B) [1−P(A)]
= P(A) + P(B). P(A′)
= 1− P(A′) + P(B) P(A′)
= 1− P(A′) [1− P(B)]
= 1− P(A′) P (B′)

Q.19. 20 cards are numbered from 1 to 20. If one card is drawn at random, what is the probability that the number on the card is a prime number?
(a) ⅕
(b) ⅖
(c) ⅗
(d) 5

Correct Answer is option (b)
Let E be the event of getting a prime number.
E = {2, 3, 5, 7, 11, 13, 17, 19}
Hence, P(E) = 8/20 = 2/5.

Q.20. The probability of solving the specific problems independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that exactly one of them solves the problem.
(a) 1
(b) ½
(c) ⅓
(d) ¼

Correct Answer is option (b)
P(A) = ½
P(B) = ⅓
Since, A and B are independent events, therefore;
⇒ P (A ∩ B) = P (A). P (B)
⇒ P (A ∩ B) = ½ × 1/3 = 1/6
P (A’) = 1 – P (A) = 1 – 1/2 = 1/2
P (B’) = 1 – P (B) = 1 – 1/3 = 2/3
Now the probability that exactly one of them solved the problem is either the problem is solved by A and not B or vice versa.
P (A).P (B’) + P (A’).P (B)
= ½ (2/3) + ½ (1/3)
= 1/3 + 1/6 = 3/6
⇒ P (A).P (B’) + P (A’).P (B) = ½