Answer:
0.050328
Step-by-step explanation:
Given that there are 10 coins, out of which 9 are fair and one is double headed.
P(selecting fair coin) = 0.9 and P(selecting double headed coin) = 0.1
P(H/fair coin)= 0.5 and P(H/doube headed ) = 1
Probability of getting in 1 toss head = Prob of selecting fair coin and getting head+ Prob of selecting double headed and getting head
= [tex]0.9*0.5+0.1*1\\=0.55[/tex]
This will be constant p for 10 trials going to be done because coin is replaced each time.
Also there are only two outcomes
X no of heads in 5 flips is Binomial with (5, 0.55)
P(x=5) = Reqd probability
= [tex]0.55^5\\=0.050328[/tex]
