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- Most Important Multiple Choice Questions
- Online Probability Exercise with Correct Answer Key and Solutions
- Useful for all Competitive Exams
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- Question 1 of 20
In a box there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
If the drawn ball is neither red nor green, then it must be blue, which can be picked in ⁷C₁ = 7 ways. One ball can be picked from the total (8 + 7 + 6 = 21) in ²¹C₁ = 21 ways.
∴ Required probability
- Question 2 of 20
A bag contains 5 green and 7 red balls. Two balls are drawn. The probability that one is green and the other is red is
There are 5 + 7 = 12 balls in the bag and out of these two balls can be drawn in ¹²C₂ ways. There are 5 green balls, therefore, one green ball can be drawn in ⁵C₁ ways; similarly, one red ball can be drawn in ⁷C₁ ways so that the number of ways in which we can draw one green ball and the other red is ⁵C₁ × ⁷C₁.
Hence, P (one green and the other red)
- Question 3 of 20
A bag contains 5 white and 7 black balls and a man draws 4 balls at random. The odds against these being all black is :
There are 7 + 5 = 12 balls in the bag and the number of ways in which 4 balls can be drawn is ¹²C₄ and the number of ways of drawing 4 black balls (out of seven) is ⁷C₄.
Hence, P (4 black balls)
Thus the odds against the event ‘all black balls’ are
- Question 4 of 20
Four balls are drawn at random from a bag containing 5 white, 4 green and 3 black balls. The probability that exactly two of them are white is :
Total number of balls = 12
Hence, required probability
No of ways of drawing 2 white balls from 5 white balls = ⁵C₂.
Also, No of ways of drawing 2 other from remaining 7 balls = ⁷C₂
- Question 5 of 20
The probability that when 12 balls are distributed among three boxes, the first will contain three balls is,
Since each ball can be put into any one of the three boxes. So, the total number of ways in which 12 balls can be put into three boxes is 3¹².
Out of 12 balls, 3 balls can be chosen in ¹²C₃ ways. Now, remaining 9 balls can be put in the remaining 2 boxes in 2⁹ ways. So, the total number or ways in which 3 balls are put in the first box and the remaining in other two boxes is ¹²C₃ × 2⁹.
Hence, required probability
- Question 6 of 20
A bag contains 2 red, 3 green and 2 blue balls. 2 balls are to be drawn randomly. What is probability that the balls drawn contain no blue ball?
2 balls can be drawn in the following ways
1 red and 1 green or 2 red or 2 green
- Question 7 of 20
A bag has 13 red, 14 green and 15 white balls, p₁ is the probability of drawing exactly 2 white balls when four balls are drawn. Then the number of balls of each colour are doubled. Let p₂ be the probability of drawing 4 white balls when 8 ball are drawn, then
⇒ p₁ > p₂
- Question 8 of 20
In a box carrying one dozen of oranges, one-third have become bad. If 3 oranges are taken out from the box at random, what is the probability that at least one orange out of the three oranges picked up is good?
No. of selection of 3 oranges out of the total 12 oranges
= ¹²C₃ = 2 × 11 × 10 = 220.
No. of selection of 3 bad oranges out of the total 4 bad oranges = ⁴c₃ = 4
∴ n(E) = no. of desired selection of oranges
= 220 – 4 = 216
∴ P (E) =
- Question 9 of 20
A box contains 5 green, 4 yellow and 3 white marbles, 3 marbles are drawn at random. What is the probability that they are not of the same colour?
Total no. of ways of drawing 3 marbles
Total no. of ways of drawing marbles, which are of same colour
= ⁵C₃ + ⁴C₃ + ³C₃
= 10 + 4 + 1 = 15
∴ Probability of same colour
= = ³⁄₄₄
∴ Probability of not same colour
= 1 – ³⁄₄₄ = ⁴¹⁄₄₄
- Question 10 of 20
Two dice are thrown simultaneously. The probability of obtaining a total score of seven is
When two are thrown then there are 6 × 6 exhaustive cases
∴ n = 36. Let A denote the event “total score of 7” when 2 dice are thrown then A
= [(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)].
Thus there are 6 favourable cases.
∴ m = 6
- Question 11 of 20
In each of a set of games it is 2 to 1 in favour of the winner of the previous game. The chance that the player who wins the first game shall win three at least of the next four is
Let W stand for the winning of a game and L for losing it. Then there are 4 mutually exclusive possibilities
(i) W, W, W
(ii) W, W, L, W
(iii) W, L, W, W
(iv) L, W, W, W.
[Note that case (i) includes both the cases whether he losses or wins the fourth game.]
By the given conditions of the question, the probabilities for (i), (ii), (iii) and (iv) respectively are
Hence the required probability
[Since the probability of winning the game if previous game was also won is 2/(1+2) = 2/3 and the probability of winning the game if previous game was a loss is 1/(1+2) = 1/3].
- Question 12 of 20
A man and his wife appear for an interview for two posts. The probability of the husband’s selection is ¹⁄₇ and that of the wife’s selection is ⅕. The probability that only one of them will be selected is
Probability that only husband is selected
Probability that only wife is selected
∴ Probability that only one of them is selected
- Question 13 of 20
In a single throw with four dice, the probability of throwing seven is:
Total of seven can be obtained in the following ways
1, 1, 1, 4 in = 4 ways [there are four objects, three repeated]
1, 1,2, 3 in = 12 ways
1, 2,2, 2 in = 4 ways
Hence, required probability
[Since exhaustive no. of cases = 6 × 6 × 6 × 6 = 6⁴]
- Question 14 of 20
Six dice are thrown. The probability that different number will turn up is :
The number of ways of getting the different number 1, 2, ….., 6 in six dice = 6!.
Total number of ways = 6⁶
Hence, required probability
- Question 15 of 20
There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is
The faulty machines can be identified in two tests only if both the tested machines are either all defective or all non-defective. See the following tree diagram.
(Here D is for Defective & ND is for Non Defective)
Since the probability that first machine is defective (or non-defective) is ²⁄₄ and the probability that second machine is also defective (or non – defective) is ⅓ as 1 defective machine remains in total three machines.
- Question 16 of 20
The probability that a person will hit a target in shooting practice is 0.3. If he shoots 10 times, the probability that he hits the target is
The probability that the person hits the target = 0.3
∴ The probability that he does not hit the target in a trial
= 1 – 0.3 = 0.7
∴ The probability that he does not hit the target in any of the ten trials = (0.7)¹⁰
∴ Probability that he hits the target
= Probability that at least one of the trials succeeds
= 1 – (0.7)¹⁰.
- Question 17 of 20
The probability that A can solve a problem is ⅔ and B can solve it is ¾. If both attempt the problem, what is the probability that the problem gets solved?
The probability that A cannot solve the problem
The probability that B cannot solve the problem
The probability that both A and B cannot solve the problem
∴ The probability that at least one of A and B can solve the problem
∴ The probability that the problem is solved = ¹¹⁄₁₂
- Question 18 of 20
Seven people seat themselves indiscriminately at round table. The probability that two distinguished persons will be next to each other is
Seven people can seat themselves at a round table in 6! ways. The number of ways in which two distinguished persons will be next to each other = 2 (5) !, Hence, the required probability
- Question 19 of 20
The probability that two integers chosen at random and their product will have the same last digit is :
The condition implies that the last digit in both the integers should be 0, 1, 5 or 6 and the probability
[ Since the squares of numbers ending in 0 or 1 or 5 or 6 also 0 or 1 or 5 or 6 respectively]
- Question 20 of 20
The probability that the birth days of six different persons will fall in exactly two calendar months is
Exhaustive number of cases = 12
Favourable cases = ¹²C₂ (2⁶ – 2)