3 years ago

If 1/6 of Aurora’s candy is strawberry & 4/6 are chocolate what is the friction of the candy in the box

1 answer:

4 0

2/3rds is candy gkgkcjvj

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Aleksandr-060686 [28]

Answer:

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alexdok [17]

Answer:

0.12 is the required probability.

Step-by-step explanation:

We are given the following in the question:

A: next component brought in for repair is an audio component

B: event that the next component is a compact disc player

$B \in A$

$P(A)= 0.6\\P(B) =0.05$

We have to evaluate:

$P(B|A) = \dfrac{P(B\cap A)}{P(A)}\\P(B|A) = \dfrac{P(B)}{P(A)}\\\\P(B|A) = \dfrac{0.05}{0.6} = 0.12$

0.12 is the required probability.

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3 years ago