We have two coins, a and b. for each toss of coin a, we obtain heads with probability 1/2 ; for each toss of coin b, we obtain heads with probability 1/3 . all tosses of the same coin are independent. we toss coin a until heads is obtained for the first time. we then toss coin b until heads is obtained for the first time with coin b. the expected value of the total number of tosses is:

The expected value is 5.

Step-by-step explanation:

Let X represent the number of tosses until the event described in the question happens. Let Y represent the number of tosses with coin A until Heads is obtained.Let Z represent the number of tosses with coin B until Heads is obtained.

As we can see, X=Y+Z. Then, by the linearity of the expected value operator, we have that

We will compute E(Y) and E(Z).

Observe that Y and Z have countable sets of outcomes (1,2,3,....) then,

,

,

Then:

for each , the probability of Y=n is given by (because the first n-1 tosses must be Tails and the n-th must be Heads). Therefore

For each , the probability of Z=n is given by (because the first n-1 tosses must be Tails and the n-th must be Heads). Therefore

Observe that, by the geometric series formula:

Therefore

Finally, E(X)=E(Y)+E(Z)=2+3=5.

You need to divide 210 square inches by 35 inches and that’s going to give you 6 inches and that’s the length of the hypothenuse of the right triangle i hope that this you : )

Idk idk idk idk idk idk idk idk