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3 years ago

Pleaseeeeeee help verrrrryyyyy confuseddddddd on #3

Mathematics

1 answer:

Dmitriy789 [7]3 years ago

5 0

X (adult)= 130
y (child)= 392
(hope you can see it...)

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Consider a situation in which P(A) = 1/8, P(C) = 1/4, and P(A and B) = 1/12. What is P(B and C)?

V125BC [204]

ANSWER

$P(B \: and \: C) = \frac{1}{6}$

EXPLANATION

Recall that,

$P(A \: and \: B)=P(A) \times P(B)$

But we were given that,

$P(A)= \frac{1}{8}$

and

$P(A \: and \: B) = \frac{1}{12}$

We substitute these values into the above formula to obtain,

$\frac{1}{12} = \frac{1}{8} \times P(B)$

This implies that,

$\frac{1}{12} \times 8=8 \times \frac{1}{8} \times P(B)$

This simplifies to,

$P(B) = \frac{8}{12}$

$P(B) = \frac{2}{3}$

So we can now find
$P(B \: and \: C).$

We use the same formula again,

$P(B \: and \: C) = P(B) \times P(C)$
We substitute the values to get,

$P(B \: and \: C) = \frac{2}{3} \times \frac{1}{4}$
We multiply out to get,

$P(B \: and \: C) = \frac{1}{6}$

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