Chapter 8.6, Problem 28E

To determine

To find: the probability for the experiment of drawing two marbles at random (without replacement ) form a bag containing one green, two yellow , and three red marbles.

Answer to Problem 28E

The probability of this event is $\frac{2}{5}$ .

Explanation of Solution

Given information: When neither marble is yellow.

Formula used:

If an event has n(E) equally likely outcomes and its sample space S has n(S) equally likely outcomes, then the probability of event is given by,

  $P(E)=\frac{n(E)}{n(S)}.$

Formula for the number of combinations of n numbers taken r numbers at a time is,

  $n(S){=}_n{C}_r.$

Calculation:

There are ${}_6{C}_2=\frac{6\times 5}{2\times 1}=15,$ different ways to choose two marbles from this bag of 6 marbles.

So, n(S) =15.

There are 2 ways to draw a green and a yellow. There are 6 ways to draw a yellow and a red. Also one could draw both yellows. So, there are 9 ways to draw at least one yellow and 6 ways to draw no yellow.

So, n (E) = 6.

Therefore, the probability of this event,

  $P(E)=\frac{n(E)}{n(S)}=\frac{6}{15}=\frac{2}{5}.$