Question

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:

Answer 1

Answer:

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

$E(X)=E(Y)+E(Z).$

• We will compute E(Y) and E(Z).

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

$E(X)=\sum^\infty_{n=1}nP(Y=n)$,

$E(Z)=\sum^\infty_{n=1}nP(Z=n)$,

Then:

• for each $n\in \mathbb{N}$, the probability of Y=n is given by $(0.5)^{n-1}(0.5)=(0.5)^{n}$ (because the first n-1 tosses must be Tails and the n-th must be Heads). Therefore

$E(Y)=\sum^\infty_{n=1}nP(Y=n)=\sum^\infty_{n=1}n(\frac{1}{2} )^n=\\\\\sum^\infty_{m=1}\sum^\infty_{n=m}(\frac{1}{2} )^n=\sum^\infty_{m=1}(\frac{1}{2} )^{m-1}=\sum^\infty_{m=0}(\frac{1}{2} )^{m}=2.$

• For each $n\in \mathbb{N}$, the probability of Z=n is given by $(\frac {2}{3})^{n-1}(\frac {1}{3})$ (because the first n-1 tosses must be Tails and the n-th must be Heads). Therefore

$E(Z)=\sum^\infty_{n=1}nP(Z=n)=\frac{1}{3}\sum^\infty_{n=1}n(\frac{2}{3} )^{n-1}=\frac{1}{3}\sum^\infty_{m=1}\sum^\infty_{n=m}(\frac{2}{3} )^{n-1}$

Observe that, by the geometric series formula:

$\sum^\infty_{n=m}(\frac{2}{3} )^{n-1}=\sum^\infty_{n=1}(\frac{2}{3} )^{n-1}-\sum^{m-1}_{n=1}(\frac{2}{3} )^{n-1}=3-\sum^{m-1}_{n=1}(\frac{2}{3} )^{n-1}=\\\\3-\sum^{m-2}_{n=0}(\frac{2}{3} )^{n}=3-\frac{1-(\frac{2}{3})^{m-1} }{1-\frac{2}{3}}=3(\frac{2}{3})^{m-1}$

Therefore

$E(Z)=\frac{1}{3}\sum^\infty_{m=1}\sum^\infty_{n=m}(\frac{2}{3} )^{n-1}=\frac{1}{3}\sum^\infty_{m=1}3(\frac{2}{3})^{m-1} =\\\\ \sum^\infty_{m=1}(\frac{2}{3})^{m-1} = \sum^\infty_{m=0}(\frac{2}{3})^{m} =3.$

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

Answer 2

Using the binomial distribution, it is found that the expected value of the total number of tosses is of 5.

For each coin, there are only two possible outcomes, either it is heads, or it is tails. The outcome of a coin is independent of any other coin, hence the binomial distribution is used to solve this question.

What is the binomial probability distribution?

It is the probability of exactly x successes on n repeated trials, with p probability of a success on each trial.

The expected number of trials until q successes is:

$E_s(X) = \frac{q}{p}$

For coin A, we obtain Heads with probability 1/2, hence:

$E_{sA}(X) = \frac{1}{\frac{1}{2}} = 2$

For coin B, we obtain Heads with probability 1/3, hence:

$E_{sB}(X) = \frac{1}{\frac{1}{3}} = 3$

2 + 3 = 5, hence, the expected value of the total number of tosses is of 5.

More can be learned about the binomial distribution at https://brainly.com/question/14424710