Question 2.31: Out of a set of 3 keys, only 1 opens a certain door. Someone......
Out of a set of 3 keys, only 1 opens a certain door. Someone tries the keys successively and let A_{ k} be the event that the right key appears the kth time. Calculate the probability P(A_{ k}):
(i) If the keys tried are not replaced, {k} = 1, 2, 3.
(ii) If the keys tried are replaced, { k} = 1, 2, . . . .
(i) ~~~P(A_{1})={\textstyle{\frac{1}{3}}};\;P(A_{2})={\textstyle{\frac{2\times1}{3\times2}}}={\textstyle{\frac{1}{3}}};\;P(A_{3})={\textstyle{\frac{2\times1\times1}{3\times2\times1}}}={\textstyle{\frac{1}{3}}}\cdot{\mathrm{So}},\;P(A_{1})=P(A_{2})=P(A_{3})=\frac{1}{3} \simeq 0.333
(ii) Clearly, P(A_{k})=P(W_{1}\cap\cdot\cdot\cdot\cap W_{k-1}\cap R_{k})=(\textstyle{\frac{2}{3}})^{k-1}\times\textstyle{\frac{1}{3}}~~\mathrm{for\ all}\,k=1,2,\cdot\cdot\cdot
REMARK 7 To calculate the probabilities in part (i) in terms of conditional probabilities, set: ~R_{k}=~ “the right key appears the kth time,” {W}_{k}=\, “a wrong key appears the kth time,” k\,=\,1,2,3. Then: P(A_{1})\,=\,P(R_{1})\,=\,{\textstyle{\frac{1}{3}}},\,P(A_{2})\,= P(W_1 \cap R_2)=P(R_2 | W_1)P(W_1)=\frac{1}{2} \times \frac {2}{3}=\frac {1}{3}, and P(A_{3})=P(W_{1}\cap {W}_{2}\cap R_{3})= P(R_{3}\mid W_{1}\cap W_{2})P(W_{2}\mid W_{1})P(W_{1})={\textstyle\frac{1}{1}}\times{\textstyle\frac{1}{2}}\times{\textstyle\frac{2}{3}}={\textstyle\frac{1}{3}}.