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galina1969 [7]
3 years ago
7

A homeowner finds that there is a 0.15 probability that a flashlight does not work when turned on. If she has three flashlights,

find the probability that at least one of them works when there is a power failure. Find the probability that the second flashlight works given that the first flashlight works.
Mathematics
1 answer:
prohojiy [21]3 years ago
6 0

Answer:

A) 0.386

B).0.325

Step-by-step explanation:

probability p will not work = 0.15

Q probability w5ill work = 0.85

Number of flashlight = 3

Probabilty of at least one=

1- probability of none

But probability of none= 3C0(p)^0(q)^3

= 1*(0.15)^0(0.85)^3

= 1*1*(0.614)

= 0.614

Probability of at least one=1-0.614

= 0.386

Probability of two flashlight work

= 3C1(0.15)^1(0.85)^2

= 3*(0.15)*(0.7225)

= 0.325

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Given

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n(S)= 36

The pairs that adds up to 4 are:

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n(Sum)  =3

So, the probability is:

Pr = \frac{n(Sum)}{n(S)}

Pr = \frac{3}{36}

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