MikkalThomas14
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Denis examined the records of the clients who had joined his gym more than six months ago, and he found these statistics: \begin{aligned} &P(\text{joined in January})=0.12 \\\\ &P(\text{member for over 6 months})=0.5 \\\\ &P(\text{over 6 months and January})=0.024 \end{aligned} ​ P(joined in January)=0.12 P(member for over 6 months)=0.5 P(over 6 months and January)=0.024 ​ Find the probability that a client remained a member for more than 666 months, given that the client joined in January.

Respuesta :

Answer:

.2

Step-by-step explanation:

This is a conditional probability question. So it's asking you to find the probability that a client remained a member for more than 6 months, given that the client joined in January - which is formatted as P = (.5 | .12). You would then divide the chance of being over 6 months AND in January over the chance of being a member for over six months. ( .024 / .12) There, you would get .2 as your answer.

The probability that a client remained a member for 6  months given that the clients joined in January is 0.2.

Given,

P(joined in January)=0.12

P(over 6 months and January)=0.024

P(over 6 months/January)=[tex]\frac{0.024}{0.12}=0.2[/tex]

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