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: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[tex]E(X)=E(Y)+E(Z).[/tex]We will compute E(Y) and E(Z). Observe that Y and Z have countable sets of outcomes (1,2,3,....) then,[tex]E(X)=\sum^\infty_{n=1}nP(Y=n)[/tex],[tex]E(Z)=\sum^\infty_{n=1}nP(Z=n)[/tex],Then: for each [tex]n\in \mathbb{N}[/tex], the probability of Y=n is given by [tex](0.5)^{n-1}(0.5)=(0.5)^{n}[/tex] (because the first n-1 tosses must be Tails and the n-th must be Heads). Therefore[tex]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.[/tex]For each [tex]n\in \mathbb{N}[/tex], the probability of Z=n is given by [tex](\frac {2}{3})^{n-1}(\frac {1}{3})[/tex] (because the first n-1 tosses must be Tails and the n-th must be Heads). Therefore[tex]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}[/tex]Observe that, by the geometric series formula:[tex]\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}[/tex]Therefore [tex]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.[/tex]Finally, E(X)=E(Y)+E(Z)=2+3=5.