Answer to Question #126320 in Statistics and Probability for Zoya

Question

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. 

Answer 1

Let A be the event that Amit has taken at least one winning lottery ticket.

And let B be the event that Amit doesn't have a winning lottery tickets.

Then AUB=U, A$\bigcap$B=$\varnothing$

and P(A)+P(B)=1

P(B)=m/n

There are 7 blank tickets.

So m=₇C₃=7!/3!/4!=35,

n=₁₀C₃=10!/3!/7!=120.

Here _rC_k=r!/k!/(r-k)! is binomial coefficient.

P(B)=35/120=7/24,

P(A)=1-P(B)=1-7/24=17/24.

Let C be the event that Somna has taken winning lottery ticket.

P(C)=m/n.

There are 2 winning tickets.

So m=2.

n=5.

P(C)=2/5

17/24=85/120>48/120=2/5 => P(A)>P(C).

Answer:

Amit has a better chance of winning a prize.