Chapter 6.5, Problem 84HP

To determine

To find: the probability that the tiles numbered 5 and 6 are the last two drawn.

Answer to Problem 84HP

  $\frac{1}{15}$ .

Explanation of Solution

Given:

Tiles numbered from 1 to 6 are placed in the bag.

Calculation:

According to the question, compute the probability of 5th and 6th tiles.

Therefore, probability of red marble is:

  $P\left(6\right)\text{=}\frac{1}{6}\text{ {as total number of tiles is 6}}$

And the probability of green marble (without replacing red marble) is:

  $P\left(5\right)\text{=}\frac{1}{5}\text{ {as total number of marble is 5}}$

Thus, the probability that the tiles numbered 5 and 6 are the last two drawn is $P\left(5\text{ and 6}\right)\text{=}\frac{1}{6}\times \frac{1}{5}+\frac{1}{5}\times \frac{1}{6}=\frac{1}{30}+\frac{1}{30}=\frac{1}{15}$ .

Hence, the probability that the tiles numbered 5 and 6 are the last two drawn is $\frac{1}{15}$ .