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FrozenT [24]
2 years ago
5

Let p(z)=0.54, p(y)=0.22, and p(z u y)= 0.62. Find each probability.

Mathematics
1 answer:
Marysya12 [62]2 years ago
3 0

Answer : p(only z)=0.40 and p(only u)=0.08

Explanation :

We have given that ,

p(u)=0.22,

p(z∪u)=0.62,

p(z)=0.54

By using probability rule of union of 2 events, we have,

p(z)+p(u)-p(z∩u)=p(z∪u)

⇒0.54+0.22-p(z∩u)=0.62

⇒0.76-p(z∩u)=0.62

⇒(-)p(z∩u)=0.62-0.76

⇒(-)p(z∩u)= (-)0.14

⇒p(z∩u)=0.14

Now,

p(only z)=p(z)-p(z∩u)

              =0.54-0.14

              =0.40

and,

p(only u)=p(u)-p(z∩u)

             =0.22-0.14

             =0.08

∴ Each probability will be

p(only z)=0.40

p(only u)=0.08

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Answer:

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Solution:

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The proportion of the given sample, p = \frac{38}{60} = 0.634

Therefore, the probability is given by:

P(p\leq 0.634) = [\frac{p - P'}{\sqrt{\frac{P'q'}{n}}}]\leq [\frac{0.634 - 0.74}{\sqrt{\frac{0.74\times 0.26}{60}}}]

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Step-by-step explanation:

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