Chapter 12, Problem 44CR

To determine

Find probability $P\left(redthenred\right)$ .

Answer to Problem 44CR

  $P\left(redthenred\right)=\frac{2}{7}$

Explanation of Solution

Given information:

You randomly pick two marbles from a bag containing $3$ yellow marbles and $4$ red marbles; you pick the second marble without replacing the first marble.

Calculation:

Consider the given data,

A bag containing $3$ yellow marbles and $4$ red marbles,

Total marbles in bag $7$ , pick the second marble without replacing the first marble.

So the number of marbles for second choice is less than one, means it affect the probability of second event, first and second events are dependent.

It is formulate as,

  $P\left(AthenB\right)=P\left(A\right)\cdot P\left(BafterA\right)$

Similarly, $P\left(redthenred\right)=P\left(red\right)\cdot P\left(redafterred\right)$ ,

  $\begin{array}{l}P\left(event\right)=\frac{numberoffavorableoutcomes}{numberofpossibleoutcomes}\\ P\left(red\right)=\frac{4}{7}\end{array}$

The second marble is picked up without replacing the first, so the number of marbles available for second choice is $6$ .

The number of possible outcomes to choose a marble out of six is six.

Out of six, three marbles are red, as one red is already chosen, without replacing.

It means the number of favorable outcomes to choose a red marble is three.

Now,

  $\begin{array}{l}P\left(redafterred\right)=\frac{3}{6}\\ =\frac{1}{2}\\ now,\\ P\left(redthenred\right)=\frac{4}{7}\cdot \frac{1}{2}\\ =\frac{2}{7}\end{array}$

Hence, $P\left(redthenred\right)=\frac{2}{7}$ .