A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the first actuarial exam in June. If she passes that exam, then she will take the second exam in July, and if she also passes that one, then she will take the third exam in September. If she fails an exam, then she is not allowed to take any others. The probability that she passes the first exam is$.9$. If she passes the first exam, then the conditional probability that she passes the second one is $.8$, and if she passes both the first and the second exams, then the conditional probability that she passes the third exam is .7.

(a) What is the probability that she passes all three exams?

(b) Given that she did not pass all three exams, what is the conditional probability that she failed the second exam?

Step by step solution

01

Step 1:Given Information(part a)

From the given data,

Probability that she breezes through the first exam $P\left(E_1\right)=0.9$

Probability that she breezes through the second exam given that she finishes first exam is, $P\left(E_2\mid E_1\right)=0.8$

Probability that she breezes through the third exam given that she finishes first and second exams is, $P\left(E_2\mid E_1\cap E_1\right)=0.7$

02

Step 2:Explanation(part a)

The probability that she passes all three exams,

$P\left(E_1\cap E_2\cap E_3\right)=P\left(E_3\right)$

$=P\left(E_1\right)\times P\left(E_2\mid E_1\right)\times P\left(E_3\mid E_1\cap E_2\right)$

$=\left(0.9\right)\left(0.8\right)\left(0.7\right)$

$=0.504$

03

Step 3:Final Answer(part a)

$0.504$is the probability that she passes all three exams.

04

Step 4:Given Information(part b)

Given that she did not pass all three exams .

05

Step 5:Explanation(part b)

The conditional probability that she failed the second exam is,

$P\left(E_1\cap E_2^c\mid E_3^c\right)=\frac{P\left(E_1\cap E_2^c\cap E_3^c\right)}{P\left(E_3^c\right)}$

$=\frac{P\left(E_1\cap E_2^C\right)}{1-P\left(E_1\cap E_2\cap E_3\right)}$

$=\frac{P\left(E_1\right)P\left(E_2^c\mid E_1\right)}{1-0.504}$

$=\frac{P\left(E_{\mathrm{y}}\right)\left(1-P\left(E_2\mid E_{\mathrm{f}}\right)\right)}{0.496}$

$=\frac{\left(0.9\right)(1-0.8)}{0.496}$

$=\frac{\left(0.9\right)\left(0.2\right)}{0.496}$

$=\frac{0.18}{0.496}$

$=0.3629$

06

Step 6:Final Answer(part b)

$0.3629$is the conditional probability that she failed the second exam .

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!