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Licemer1 [7]
2 years ago
12

A doctor gives a patient a 6060​% chance of surviving bypass surgery after a heart attack. if the patient survives the​ surgery,

then the patient has a 5050​% chance that the heart damage will heal. find the probability that the patient survives the surgery and the heart damage heals.
Mathematics
1 answer:
DENIUS [597]2 years ago
3 0
Let BS be the event that the patient survives bypass surgery. Let H be the event that the heart damage will heal. Then P(BS) = 0.60, and also we have a conditional probability: GIVEN that the patient survives, the probability that the heart damage will heal is 0.5, that is P(H|BS) = 0.5 We want to know P(BS and H). Using the formula of the conditional probability: P(H and BS) = P(H|BS)·P(BS) = (0.6)(0.5) = 0.3 
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romanna [79]

Answer:

0.9936 = 99.36% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

Step-by-step explanation:

For each disk, there are only two possible outcomes. Either it works, or it does not. The probability of a disk working is independent of any other disk, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Assume that there is a 8​% rate of disk drive failure in a year.

So 100 - 8 = 92% probability of working, which means that p = 0.92

Two disks are used:

This means that n = 2

What is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive?

This is:

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{2,0}.(0.92)^{0}.(0.08)^{2} = 0.0064

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0064 = 0.9936

0.9936 = 99.36% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

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