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Ede4ka [16]
3 years ago
15

A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the fir

st 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?
Mathematics
1 answer:
erica [24]3 years ago
5 0

Answer:

(a) 0.7

(b) \frac 13

Step-by-step explanation:

Let E_1, E_2, and E_3 be the events of passing the 1st, 2nd, and 3rd exam individually.

Given that P(E_1)=0.9\;\cdots(i)

where P(E_1) denotes the probability of passing the 1st exam.

Condition for passing the second exam is, at first, the candidate must have to pass the 1st exam.

So, P\left(\frac{E_2}{E_1}\right)=0.8\;\cdots(ii)

where P(E_2/E_1) denotes the probability of passing the 2nd exam when she already passed the 1st exam (given, 0.8).

Similarly, as the conditional probability of passing the 3rd exam is 0.7, and the condition for this is, at first, she must have to pass the 1st and 2nd exam. i.e,

P\left(\frac{E_2}{P(E_2/E_1)}\right)=0.7\;\cdots(iii)

(a) For passing all the exams, the condition is, at first, she has to pass the 1st and 2nd exam, then she has to pass the 3rd exam too. The probability for this conditional has been given as 0.7.

So, the probability that she passes all three exams is 0.7.

(b) Given that she didn't pass all three exams that means she either failed in 1st exam or she passed the 1st and failed in 2nd exam or she passed both 1st and 2nd but failed in the 3rd exam.

Let F be the event that she didn't pass all three exams. So,

P(F)=(1-P(E_1))+\left(1-\frac{P(E_2)}{P(E_1)}\right)+\left(1-\frac{P(E_2)}{P(E_2/E_1)}}\right)

\Rightarrow P(F)=(1-0.9)+(1-0.8)+(1-0.7)=0.6

Lef F_2 be the event that she failed the 2nd exam, so

P(F_2)=1-\frac{P(E_2)}{P(E_1)}

\Rightarrow P(F_2)=1-0.8=0.2

So, the conditional probability that she failed the 2nd exam is

P\left(\frac{F_2}{F}\right)=\frac{P(F_2)}{P(F)}

\Rightarrow P\left(\frc{F_2}{F}\right)=\frac{0.2}{0.6}=\frac{1}{3}=0.33

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The probability that a randomly chosen plate contains the number 2222 is 0.000028 approximately.

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Part (A)

Let A is the ways in which plates consist of three letters followed by three digits and B is the ways in which two letters followed by four digits.

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