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Juan selects 2 colored pencils randomly There are 4 groen, 3 yellow, 4 red, 3 blue and 2 orange What is the probability Juan chooses a yellow pencil putsit back then chooses a red pencil?7/3212/327/2563/64

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There are 16 pencils

4 are green

3 are yellow

4 are red

3 are blue

2 are orange

To determine the probability of randomly choosing a yellow pencil and a red pencil, you have to apply the theorem of the product of probabilities:


P(Y\cap R)=P(Y)\cdot P(R)

Since Juan puts back the first pencil before taking the second one, both events "yellow" and "red" are independent.

The individual probabilities for both events can be calculated as the quotient between the number of pencils of each color and the total number of pencils:


\begin{gathered} P(Y)=\frac{nº\text{yellow}}{\text{Total}} \\ P(Y)=(3)/(16) \end{gathered}
\begin{gathered} P(R)=\frac{nº\text{red}}{\text{total}} \\ P(R)=(4)/(16) \\ P(R)=(1)/(4) \end{gathered}

Then the probability of selecting a yellow pencil and then a red pencil can be determined as follows:


\begin{gathered} P(Y\cap R)=P(Y)\cdot P(R)=(3)/(16)\cdot(1)/(4)=(3\cdot1)/(16\cdot4) \\ P(Y\cap R)=(3)/(64) \end{gathered}

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