Question

three mothers labeled 12 and three give birth in a hospital the nurses after feeding the babies return the babies in random order.

what is the probability that exactly one of the mothers got the right baby?

what is the probability that exactly to the mothers got the right baby?

what is the probability that all three of the mothers got the right baby?

what is the likelihood that at least one of the mothers got the right baby?

what is the expected value of the number of matches?

Answer 1

let M1,M2,M3 denotes mother 1,mother 2 and mother 3

B1,B2,B3 denotes baby from mother 1,mother 2 and mother 3 respectively

so there are 3 possible pairs of mothers and Babies that are given below

{(M1,B1),(M2,B2),(M3,B3),(M1,B2),(M1,B3),(M2,B1),(M2,B3),(M3,B1),(M3,B2)}

while possible combinations of pairs is

{(M1B1,M2B2,M3B3),(M1B1,M2B3,M3B2),(M2B2,M1B3,M3B1),(M3B3,M1B2,M2B1),(M1B2,M2B3,M3B1),(M1B3,M2B1,M3B2)}

let X is number of mathces

1)

For exactly 1 match there is only one match while the other two have not matched so possible outcomes are

so out of 6 possible combinations there are 3 pairs with exactly one match

these combinations are M1B1,M2B3,M3B2),(M2B2,M1B3,M3B1),(M3B3,M1B2,M2B1)

Hence required probability is P(X=1)= 3/6=0.5

2)

we have to find the probability that exactly two mothers got right baby but there are no such combinations out of hence

P(X=2)=0/6=0

3)

since there is only one combination with all three matched pairs hence

P(X=3) =1/6

4)

P(X> 1) =P(X=1)+P(X=2)+P(X=3) =(3/6)+(0/6)+(1/6) =4/6

5)

P(X=0) =1-P(X>1) =1-(4/6) =2/6

now

E(X) =0*P(X=0)+1*P(X=1)+2*P(X=2)+3*P(X=3)

=0*(2/6)+1*(3/6)+2*0+3*(1/6)

=1

Hence expected mathces is 1