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.