Solved Subjective Question on Probability Set 2

Example.1

An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the number obtained by adding the numbers on the two faces is noted. If the result is a tail, a card from a well-shuffled pack of eleven cards numbered 2, 3, 4,L, 12 is picked and the number on the card is noted. Find the probability that noted number is either 7 or 8.

Solution

Let A₁(A₂) be the event that tossing of coin results in head (tail).

And A be the event that noted number is 7 or 8.

P(A₁) = P(A₂) =

P(A/A₂) =

P(A/A₁) =

(As 7 can result in, (1, 6), (2,5), (3,4), (6,1), (5,2), (4, 3) i.e. 6 ways and 8 can result in (2, 6), (3,5), (4,4), (6,2), (5,3) i.e. 5 ways).

Now P(A) = P(A₁)×P(A/A₁) + P(A₂)×P(A/A₂)

= =

Example.2

In a multiple-choice question there are four alternative answers, of which one or more than one are correct. A candidate will get marks on the question only if he ticks all the correct answers. The candidate decides to tick answers randomly. If he is allowed up to three chances to answer the questions, find the probability that he will get marks on it.

Solution

Total number of ways of answering the question,

= ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄ = 15

Now out of these, only one combination is correct.

&⇒ Probability of answering the question correctly at the first trial =

Now, Probability of answering the question correctly at the second trial =

(with the assumption that student doesn’t repeat the mistake of first trial)

Similarly, probability of answering the question correctly at the third trial

&⇒ Probability of required event =

Example.3

Sixteen players S₁, S₂,...,S₁₆ play in a tournament. They are divided into eight pairs at random. From each pair a winner is decided on the basis of a game played between the two players of the pair. Assume that all the players are of equal strength .

(a) find the probability that the player S₁ is among the eight winner

(b) find the probability that exactly one of the two players S₁ and S₂ is among the eight winner. .

Solution

(a) Since all the players are of equal strength and eight players are winners. So probability that S₁ will be among the winner equals to 8/16 = 1/2.

(b) Let A₁ be the event that S₁, S₂ play in same pair and A₂be the event that they play in different pair and A be the event that exactly one out of S₁ and S₂ is among the winners.

P(A) = P(A₁)×P(A/A₁) + P(A₂)×P(A/A₂)

Now P(A₁) =

P(A₂) = 1 -

Therefore, P(A) = .

Example.4

A manufacturing company uses an acceptance scheme on production items before they are shipped. The plan is a two-stage one. Boxes of 25 items are readied for shipment and a sample of 3 are tested for defectives. If any defectives are found, the entire box is sent for 100% screening. If no defectives are found, the box is shipped.

(a) what is the probability that a box containing 3 defectives will be shipped?

(b) what is the probability that a box containing only one defective will be sent back for screening?

Solution

(a) A box containing 3 defectives will be shipped, if no defective item is included in the sample.

&⇒ Required probability =

(b) In this case sample taken must contain the defective item,

&⇒ Required probability =

Example.5

One of ten keys opens the door. If we try the keys one after another, what is the probability that the door is opened on the (i) first attempt, (ii) second attempt, (iii) tenth attempt.

Solution

Since out of 10 keys only one can open the door so door will open in first attempt with probability Now if he fails in first attempt which has the probability then he will attempt next time with different key. So probability that door will open in second attempt is

Probability that door will open only in tenth attempt is equal to the probability that door not opened in first nine attempts which is equal to (As each time he tries with new key and keeps away the key which does not open the door).