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DEPENDENT EVENTS AND CONDITIONAL PROBABILITY A bucket contains 12 balls, 8 of which are white. If you reach in and randomly grab two balls one at a time without replacement, what is the probability that both are white? •Enter your answer as a fraction in lowest terms.

User Doesdos
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Answer:

The probability that both are white is;


(14)/(33)

Step-by-step explanation:

We want to find the probability that both balls removed without replacement are white.


P=P_1* P_2

Where;


\begin{gathered} P_1=\text{ probability that the first ball is white} \\ P_2=\text{ Probability that the second ball is white} \end{gathered}

Solving for the first;


\begin{gathered} P_1=\frac{\text{Number of white balls}}{\text{Total number of balls}}=(8)/(12) \\ P_1=(2)/(3) \end{gathered}

For the second selection, since there is no replacement, the number of white balls would have reduced by one to 7 and the total number of balls will also reduce to 11.

So we have;


P_2=(7)/(11)

So;


\begin{gathered} P=P_1* P_2 \\ P=(2)/(3)*(7)/(11) \\ P=(14)/(33) \end{gathered}

Therefore, the probability that both are white is;


(14)/(33)

User Truthseeker
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