Answer to Question #126295 in Statistics and Probability for mishal

Question #126295

Amit has taken 3 tickets randomly from a pack of 10 lottery tickets in which 3 are winning
tickets and 7 are blanks. Somna has taken one ticket from a pack of 5 lottery tickets in which 2
are winnings and 3 blank tickets. Amongst Amit and Somna who has a better chance of winning
a prize.

Expert's answer

Probability of Amit winning the prize -

As Amit has 3 tickets in total from a pack of 10 lottery tickets in which 3 are winning tickets and 7 are blanks, he can win the prize if he can find at least one winning ticket. Conversely , if all the 3 tickets are blank , only then he loses the prize.

Therefore, probability of the first ticket to be a blank ticket is 7/10.

Probability of the second ticket to be also a blank ticked when drawn in succession is (7-1)/(10-9), as a blank ticket is already drawn before. It is equal to 6/9.

Probability of the third ticket drawn in succession to the first and second tickets to be also a blank ticket is, (6-1)/(9-1). It is equal to 5/8.

In order for Amit to not win the prize, all the three events should occur. Hence, the final probability of Amit losing the prize is - (7/10)*(6/9)*(5/8) , which is equal to 7/24 after simplification.

Hence, the probability of Amit winning the prize is only when he does not lose. That is , 1 - probability of losing , which is equal to 1 - 7/24 = 17/24.

Probability of Somna winning the prize -

Somna has one ticket from a pack of 5 lottery tickets in which 2 are winning ticket and 3 blank tickets.

Probability of Somna winning the prize is when she has the winning ticket with her.

Hence, the probability of the ticket to be a winning ticket is 2/5.

Comparing both the obtained probabilities, 17/24 (Amit) and 2/5 (Somna).

(17/24) - (2/5) = [ (17*5) - (2*24) ] / [ (24*5) ]

= [ 85 - 48 ] / [ 120 ]

= 37 / 120 which is > 0.

As, 17/24 > 2/5 , hence Amit's probablity of winning a prize is higher than that of Somna.