Answer:
Step-by-step explanation:
Given:
we have 10 coins
nine of them are fair coins, which means: [tex]P(H)=\frac{1}{2}; P(T)=\frac{1}{2}[/tex]
but one coin is biased it has head on both sides which means [tex]P(H)=\frac{1}{1}; P(T)=\frac{0}{1}=0[/tex]
expected number of heads for tossing 9 coins = [tex]9(\frac{1}{2})=\frac{9}{2}=4.5[/tex]
The coin 10 is biased with only head on both sides
The expected number of heads tossing this coin is = 1
Therefore, the expected number of heads if all 10 coins are tossed together is =[tex]4.5+1=5.5[/tex]
Using indicator random variables:
The number of unbiased coins = 9: n=9
[tex]P(H)=\frac{1}{2}=0.5\\\\F(x)=nP=9\times 0.5=9.5[/tex]
One coin is biased with only head. Therefore:
[tex]F(x)=1[/tex]
Finally, [tex]F(x)=4.5+1=5.5[/tex]
The expected number of heads = 5.5
