Consider Example 2a, but now suppose that when the key is in a certain pocket, there is a 10 percent chance that a search of that pocket will not find the key. Let R and L be, respectively, the events that the key is in the right-hand pocket of the jacket and that it is in the lefthand

pocket. Also, let SR be the event that a search of the right-hand jacket pocket will be successful in finding the key, and let UL be the event that a search of the lefthand jacket pocket will be unsuccessful and, thus, not find the key. Find P(SR|UL), the conditional probability that a search of the right-hand pocket will find the key given that a search of the left-hand pocket did not, by

(a) using the identity P(SR|UL) = P(SRUL)/P(UL) determining P(SRUL) by conditioning on whether or not the key is in the right-hand pocket, and determining P(UL) by conditioning on whether or not the key is in the lefthand pocket.

(b) using the identity P(SR|UL) = P(SR|RUL)P(R|UL) + P(SR|RcUL)P(Rc|UL).

Step by step solution

01

Given Information

Rand L are the events that the key is in the right-hand pocket of the jacket and that it is in the left-hand pocket. Also, $S_R$be the event that a search of the right-hand jacket pocket will be successful in finding the key, and let $U_L$be the event that a search of the left-hand jacket pocket will be unsuccessful and, thus, not find the key.

02

Part(a)

$$\begin{gathered} P\left(S_R\right|U_L)=\frac{P\left(S_R{U}_L\right)}{P\left(U_L\right)}=\frac{P\left(S_R\right)}{P\left(U_L\right)} \\ P(S_R\left|U_L\right)=\frac{0.45}{0.55} \\ P\left(S_R\right|U_L)=\frac{9}{11} \end{gathered}$$

03

Part (b)

This statement is same as in part (a) and can be proved in the similar manner.

$$\begin{gathered} \mathrm{Taking}\mathrm{RHS}\mathrm{of}\mathrm{above}\mathrm{equality} \\ =\frac{\mathrm{P}({\mathrm{S}}_{\mathrm{R}}\cap \mathrm{R}\cap {\mathrm{U}}_{\mathrm{L}})}{\mathrm{P}(\mathrm{R}\cap {\mathrm{U}}_{\mathrm{L}})}*\frac{\mathrm{P}(\mathrm{R}\cap {\mathrm{U}}_{\mathrm{L}})}{\mathrm{P}\left({\mathrm{U}}_{\mathrm{L}}\right)}+\frac{\mathrm{P}({\mathrm{S}}_{\mathrm{R}}\cap \overline{\mathrm{R}}\cap {\mathrm{U}}_{\mathrm{L}})}{\mathrm{P}(\overline{\mathrm{R}}\cap {\mathrm{U}}_{\mathrm{L}})}*\frac{\mathrm{P}(\overline{\mathrm{R}}\cap {\mathrm{U}}_{\mathrm{L}})}{\mathrm{P}\left({\mathrm{U}}_{\mathrm{L}}\right)} \\ =\frac{\mathrm{P}({\mathrm{S}}_{\mathrm{R}}\cap \mathrm{R}\cap {\mathrm{U}}_{\mathrm{L}})}{\mathrm{P}\left({\mathrm{U}}_{\mathrm{L}}\right)}+\frac{\mathrm{P}({\mathrm{S}}_{\mathrm{R}}\cap \overline{\mathrm{R}}\cap {\mathrm{U}}_{\mathrm{L}})}{\mathrm{P}\left({\mathrm{U}}_{\mathrm{L}}\right)} \\ =\frac{\mathrm{P}({\mathrm{S}}_{\mathrm{R}}\cap {\mathrm{U}}_{\mathrm{L}})}{\mathrm{P}\left({\mathrm{U}}_{\mathrm{L}}\right)} \\ =\mathrm{LHS}=\frac{9}{11} \end{gathered}$$

which completes the proof.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!