Question

Suppose you have a bag of 10 coins. Nine of them are fair coins, that is, if you toss any of these 9 coins the probability of getting a head, P(H) = 1/2. Similarly, probability of getting a tail, P(T) = 1/2. The other coin is biased — it has head on both sides. Use indicator random variables to compute expected number of heads, if all 10 coins are tossed together.

Answer 1

Answer:

Step-by-step explanation:

Given:

we have 10 coins

nine of them are fair coins, which means: $P(H)=\frac{1}{2}; P(T)=\frac{1}{2}$

but one coin is biased it has head on both sides which means $P(H)=\frac{1}{1}; P(T)=\frac{0}{1}=0$

expected number of heads for tossing 9 coins = $9(\frac{1}{2})=\frac{9}{2}=4.5$

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 =$4.5+1=5.5$

Using indicator random variables:

The number of unbiased coins = 9: n=9

$P(H)=\frac{1}{2}=0.5\\\\F(x)=nP=9\times 0.5=9.5$

One coin is biased with only head. Therefore:

$F(x)=1$

Finally, $F(x)=4.5+1=5.5$

The expected number of heads = 5.5