Test Preparation on Permutation, Combination, and Probability For Competitive Exam

1.

The probability that a number selected at random from the first 50 natural numbers is a composite number is -.

The number of exhaustive events = ⁵⁰C₁ = 50.
We have 15 primes from 1 to 50.
Number of favourable cases are 34.
Required probability = 34/50 = 17/25.

2.

A coin is tossed live times. What is the probability that there is at the least one tail?

Let P(T) be the probability of getting least one tail when the coin is tossed five times. = There is not even a single tail. i.e. all the outcomes are heads. = 1/32 ; P(T) = 1 - 1/32 = 31/32

3.

If a number is chosen at random from the set {1, 2, 3, ...., 100}, then the probability that the chosen number is a perfect cube is -.

We have 1, 8, 27 and 64 as perfect cubes from 1 to 100. Thus, the probability of picking a perfect cube is 4/100 = 1/25.

4.

Out of first 20 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is -.

n(S) = 20
n(Even no) = 10 = n(E)
n(Prime no) = 8 = n(P)
P(EᴜP) = 10/20 + 8/20 - 1/20 = 17/20

5.

If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is

The number of exhaustive outcomes is 36. Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2 P(E) = 1 - 1/2 = 1/2.

6.

If four coins are tossed, the probability of getting two heads and two tails is -.

Since four coins are tossed, sample space = 24 Getting two heads and two tails can happen in six ways. n(E) = six ways p(E) = 6/24 = 3/8

7.

If a card is drawn from a well shuffled pack of cards, the probability of drawing a spade or a king is -.

P(SᴜK) = P(S) + P(K) - P(S∩K), where S denotes spade and K denotes king.
P(SᴜK) = 13/52 + 4/52 - 1/52 = 4/13

8.

If six persons sit in a row, then the probability that three particular persons are always together is -

Six persons can be arranged in a row in 6! ways. Treat the three persons to sit together as one unit then there four persons and they can be arranged in 4! ways. Again three persons can be arranged among them selves in 3! ways. Favourable outcomes = 3!4! Required probability = 3!4!/6! = 1/5

9.

A bag contains 7 green and 8 white balls. If two balls are drawn simultaneously, the probability that both are of the same colour is -.

Drawing two balls of same color from seven green balls can be done in ⁷C₂ ways.
Similarly from eight white balls two can be drawn in ⁸C₂ ways.
P = ⁷C₂/¹⁵C₂ + ⁸C₂/¹⁵C₂ = 7/15

10.

A basket has 5 apples and 4 oranges. Three fruits are picked at random. The probability that at least 2 apples are picked is -.

Total fruits = 9
Since there must be at least two apples,
(⁵C₂ * ⁴C₁)/⁹C₃ + ⁵C₃/⁹C₃ = 25/42.


Right Ans	Wrong Ans	Not Attempted