Mathematics MCQ on Probability for JEE and Engineering Exam 2022
MCQ on Probability: To be an expert in JEE Mathematics, it is absolutely necessary to practice and be familiar will all the concepts as well as the questions of different types. This is essential to gain mastery over the subject. We have also often heard the common saying, “Practice Makes a Man Perfect”, hence students have to practice, practice and practice till they master the subject.
In this post we are providing you MCQ on Probability, which will be beneficial for you in upcoming JEE and Engineering Exams.
MCQ on Probability
Q1. In a bag, there are three tickets numbered 1, 2, 3. A ticket is drawn at random and put back and this is done four times. The probability that the sum of the numbers is even, is __.
a) 40/81
b) 41/81
c) 41/80
d) 40/88
Solution:
The total number of ways of selecting 4 tickets = 3^{4} = 81.
The favourable number of ways = sum of coefficients of x^{2}, x^{4},……. in (x + x^{2} + x^{3})^{4}
= sum of coefficients of x^{2}, x^{4},……. in x^{4} (1 + x + x^{2})^{4}.
Let (1 + x + x^{2})^{4 }= 1 + a_{1}x + a_{2}x^{2 }+…..+ a_{8}x^{8}.
Then 3^{4 }= 1 + a_{1} + a_{2} +…..+ a_{8},
On putting x = 1
1 = 1 + a_{1} + a_{2} +…..+ a_{8},
On putting x = −1
3^{4 }+ 1 = 2 [1 + a_{1} + a_{2} + a_{4}+ a_{6} + a_{8}]
⇒ a_{2} + a_{4}+ a_{6} + a_{8 }= 41
Thus sum of the coefficients of x^{2}, x^{4},…… = 41
Hence the required probability = 41 / 81.
Q2. If a is an integer lying in [−5, 30], then the probability that the graph y = x^{2 }+ 2 (a + 4) x − 5a + 64 is strictly above the x-axis is __________.
a) 1/3
b) 2/9
c) 4/6
d) 4/9
Solution:
x^{2 }+ 2 (a + 4) x − 5a + 64 ≥ 0
If D ≤ 0, then (a + 4)^{2} − (−5a + 64) < 0 Or
a^{2 }+ 13a − 48 < 0 Or
(a + 16) (a − 3) < 0
⇒ −16 < a < 3 ⇔ −5 ≤ a ≤ 2
Then, the favourable cases are equal to the number of integers in the interval [−5, 2], i.e., 8.
The total number of cases is equal to the number of integers in the interval [−5, 30], i.e., 36.
Hence, the required probability is 8 / 36 = 2 / 9.
Q3. 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) 20/595
b) 16/595
c) 18/595
d) 11/595
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.
Q4. 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) 5/256
b) 10/244
c) 11/243
d) 12/224
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.
Q5. If the probability of X to fail in the examination is 0.3 and that for Y is 0.2, then the probability that either X or Y failing in the examination is ______.
a) 0.55
b) 0.44
c) 0.61
d)0. 75
Solution:
Here P (X) = 0.3; P (Y) = 0.2
Now P (X ∪ Y) = P (X) + P (Y) − P (X ∩ Y)
Since these are independent events, so P (X ∩ Y) = P (X) * P (Y)
Thus required probability = 0.3 + 0.2 − 0.06 = 0.44
Q6. If any four numbers are selected and they are multiplied, then the probability that the last digit will be 1, 3, 5 or 7 is ___________.
a) 16/25
b) 15/24
c) 3/5
d) 4/21
Solution:
The total number of digits in any number at the units place is 10.
Therefore, n (S) = 10
If the last digit is 1, 3, 5 or 7, then it is necessary that the last digit in each number must be 1, 3, 5 or 7.
Therefore, n (A) = 4
P (A) = 4 / 10 = 2 / 5
Hence, the required probability is (2 / 5)^{4 }= 16 / 625.
Q7. 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) 15/13
b) 13/15
c) 12/15
d) 15/12
Solution:
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
Q8. In a box of 10 electric bulbs, two are defective. Two bulbs are selected at random one after the other from the box. The first bulb after selection is being put back in the box before making the second selection. The probability that both the bulbs are without defect is ______.
a) 4/9
b) 16/22
c) 4/5
d) 16/25
Solution:
Here P (without defect) = 8 / 10 = 4 / 5 = p
P (defected) = 2 / 10 = 1 / 5 = q and n = 2, r = 2
Hence required probability = ^{n}C_{r} * p^{r} * q^{n−r}
= ^{2}C_{2}* (45)^{2} * (15)^{0}
= 16 / 25
Q9. Forty teams play a tournament. Each team plays with every other team just once. Each game results in a win for one team. If each team has a 50% chance of winning each game, the probability that at the end of the tournament, every team has won a different number of games is _______.
a) 30! / 2^{780}.
b) 40! / 2^{750}.
c) 40! / 2^{780}.
d) 40! / 4^{780}.
Solution:
Team totals must be 0, 1, 2, 39.
Let the teams be T1, T2,……., T40, so that T1 loses to T1 for i < j. In other words, this order uniquely determines the result of every game. There are 40! such orders and 780 games, so 2^{780} possible outcomes for the games.
Hence, the probability is 40! / 2^{780}.
Q10. One coin is thrown 100 times. What is the probability of getting a tail as an odd number?
a) 5/3
b) 4/6
c) 1/2
d) 4/5
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.
Recommendation: You can also use an online probability calculator to solve the above problems of probability.
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