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olga nikolaevna [1]
2 years ago
8

Samples of laboratory glass are in small, lightpackaging or heavy, large packaging. Suppose that 2% and 1% of thesample shipped

in small and large packages, respectively, breakduring transit. (a) If 60% of the samples are shipped in largepackages and 40% are shipped in small packages, what proportion ofsamples break during shipment? (b) Also, if a sample breaks duringshipment, what is the probability that it was shipped in a smallpackage?
Mathematics
1 answer:
Aleks [24]2 years ago
5 0

Answer:

a) 1.4% of the samples break during shipment

b) the probability is 4/7 ( 57.14%)

Step-by-step explanation:

a) defining the event B= the sample of laboratory glass breaks , then the probability is:

P(B)= probability that sample is shipped in small packaging * probability that the sample breaks given that was shipped in small packaging +  probability that sample is shipped in large packaging * probability that the sample breaks given that was shipped in large packaging = 0.40* 0.02 + 0.60*0.01 = 0.014

b) we can use the theorem of Bayes for conditional probability. Then defining the event S= the sample is shipped in small packaging . Thus we have

P(S/B)= P(S∩B)/P(B) = 0.40* 0.02 / 0.014= 4/7 ( 57.14%)

where

P(S∩B)= probability that sample is shipped in small packaging and it breaks

P(S/B)= probability that sample was shipped in small packaging given that is broken

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Answer:

Step-by-step explanation:

Hello!

Given the probabilities:

P(A₁)= 0.35

P(A₂)= 0.50

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a)

Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)

Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.

b)

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c)

The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:

P(B)= (A₁∩B) + P(A₂∩B)=  0.07 + 0.025= 0.095

d)

The Bayes' theorem states that:

P(Ai/B)= \frac{P(B/Ai)*P(A)}{P(B)}

Then:

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I hope it helps!

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3 years ago
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