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JEE Maths Quiz on Probability
JEE Maths Quiz on Probability : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you Maths quiz on Probability
Maths Quiz on Probability
Q1. The probability that a man can hit a target is 3/4$ .$ He tries 5 times. The probability that he will hit the target at least three times is
a) 459/512
b) 471/ 502
c) 272 / 464
d) 239/412
a) 459/512
Q2. Two cards are drawn successively with replacement from a well shuffled deck of 52 cards then the mean of the number of aces is
a) 1/13
b) 2/13
c) 3 /13
d) 4/13
b) 2/13
Q3. A fair die is tossed repeatedly. A wins if it is 1 or 2 on two consecutive tosses and B wins if it is 3, 4, 5 or 6 on two consecutive tosses. The probability that A wins if the die is tossed indefinitely is
a) 1/3
b) 5/21
c) 16/21
d) 2/5
b) 5/21
Q4. One card is drawn randomly from a pack of 52 cards, then what is the probability that it is a king or spade?
a) 9/13
b) 2/3
c) 4/13
d) 7/15
c) 4/13
Q5. The probability that a man will be alive in 20 years is 3 / 5 and the probability that his wife will be alive in 20 years is 2 / 3. Then what is the probability that at least one will be alive in 20 years?
a) 7 / 15
b) 13 / 15
c) 13 / 17
d) 7 / 13
b) 13 / 15
Let A be the event that the husband will be alive in 20 years and B be the event that the wife will be alive in 20 years. Clearly A and B are independent events.
P (A ∩ B) = P (A) * P (B)
Given P (A) = 3 / 5, P (B) = 2 / 3
The probability that at least of them will be alive in 20 years is
P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = P (A) + P (B) − P (A) P (B)
= 3 / 5 + 2 / 3 − [3 / 5] * [2 / 3]
= [9 + 10 − 6] / [15]
= 13 / 15
Q6. If A and B are two independent events such that P (A ∩ B′) = 3 / 25 and P (A′ ∩ B) = 8 / 25, then P (A) = _______.
a) 1/5 and 3/5
b) 2/5 and 4/5
c) 3/5 and 4/7
d) None of these
a) 1/5 and 3/5
Since events are independent.
So, P (A ∩ B′) = P (A) × P (B′) = 3 / 25
⇒ P (A) × {1 − 2 P (B)} = 3 / 25 ….(i)
Similarly, P (B) × {1 − P (A)} = 8 / 25 ….(ii)
On solving (i) and (ii), we get
P (A) = 1/5 and 3/5
Q7. The probabilities that A and B will die within a year are p and q respectively, then the probability that only one of them will be alive at the end of the year is __________.
a) p – q − 4pq
b) – p + q − 4pq
c) p – q + 2pq
d) p + q − 2pq
d) p + q − 2pq
Solution:
Required probability is P [(A will die and B alive) or (B will die and A alive)]
= P [(A ∩ B′) ∪ (B ∩ A′)]
Since events are independent, so
Required probability= P (A) * P (B′) + P (B) * P (A′)
= p * (1 − q) + q * (1 − p)
= p + q − 2pq
Q8. Three ships A, B and C sail from England to India. If the ratio of their arriving safely are 2:5, 3:7 and 6:11 respectively then the probability of all the ships arriving safely is ________.
a) 18 / 595
b) 18 / 695
c) 44 / 595
d) 28 / 595
a) 18 / 595
Solution:
We have the ratio of the ships A, B and C for arriving safely to be 2:5, 3:7 and 6:11 respectively. The probability of ship A for arriving safely = 2 / [2 + 5] = 2 / 7
Similarly, for B = 3 / [3 + 7] = 3 / 10 and for C = 6 / [6 + 11] = 6 / 17 Probability of all the ships for arriving safely = 2 / 7 × 3 / 10 × 6 / 17 = 18 / 595.
Q9. One coin is thrown 100 times. What is the probability of getting a tail as an odd number?
a) 1/4
b) 3/4
c) 1/3
d) 1/2
d) 1/2
Solution:
Let p = Probability of getting tail = 1 / 2
q = Probability of getting head = 1 / 2
Also, p + q = 1 and n = 100
Required probability = P (X = 1) + P (X = 3) +….. + P (X = 99)
= ^{100}C_{1} * p * q^{99} + ^{100}C_{3} * p^{3} * q^{97} +……..+ ^{100}C_{99 }* p^{99} * q^{1}
= [(p + q)^{100 }− (p − q)^{100}] / [2]
= 1 / 2.
Q10. A bag contains 2 white and 4 black balls. A ball is drawn 5 times with replacement. The probability that at least 4 of the balls drawn are white is ________.
a) 11 / 81
b) 11 / 243
c) 11 / 27
d) 11 / 729
b) 11 / 243
Solution:
Probability for white ball = 2 / 6 = 1 / 3
Probability for black ball = 4 / 6 = 2 / 3
Required probability = ^{5}C_{5} * (1 / 3)^{5} * (2 / 3)^{0} + ^{5}C_{4} * (1 / 3)^{4} * (2 / 3)
= (1/ 3)^{4 }[(1 / 3) + [5] * [2 / 3]]
= 11 / 3^{5}
= 11 / 243
JEE Maths Quiz on Probability Based on Previous Year Paper
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