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.

Step-by-step explanation:

Given:

we have 10 coins

nine of them are fair coins, which means:

but one coin is biased it has head on both sides which means

expected number of heads for tossing 9 coins =

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 =

Using indicator random variables:

The number of unbiased coins = 9: n=9

One coin is biased with only head. Therefore:

Finally,

The expected number of heads = 5.5

i think c. but don't trust it.

step-by-step explanation:

b.-4 units

all you   do is count how many times you move