Question

(a) (i) Find the probability of getting at least one 3 when 9 fair dice are thrown. (ii) When n fair dice are thrown, the probability of getting at least one 3 is greater than 0.9. Find the smallest possible value of n. (b) A bag contains 5 green balls and 3 yellow balls. Ronnie and Julie play a game in which they take turns to draw a ball from the bag at random without replacement. The winner of the game is the first person to draw a yellow ball. Julie draws the first ball. Find the probability that Ronnie wins the game.

Answer 1

Answer:

(a)

(i) The probability of getting at least one 3 when 9 fair dice are thrown is 0.8062.

(ii) The value of n is 12.

(b) The probability that Ronnie wins the game is 0.3572.

Step-by-step explanation:

(a)

(i)

The probability of getting a 3 on a single die roll is, $p=\frac{1}{6}$.

It is provided that n = 9 fair dice are thrown together.

The outcomes of each die is independent of the others.

The random variable X can be defined as the number of die with outcome as 3.

The random variable X follows a Binomial distribution with parameters n = 9 and $p=\frac{1}{6}$.

Compute the probability of getting at least one 3 as follows:

$P(X\geq 1)=1-P(X=0)$

                $=1-[{9\choose 0}(\frac{1}{6})^{0}(1-\frac{1}{6})^{9-0}]\\\\=1-(\frac{5}{6})^{9}\\\\=1-0.19381\\\\=0.80619\\\\\approx 0.8062$

Thus, the probability of getting at least one 3 when 9 fair dice are thrown is 0.8062.

(ii)

It is provided that:

P (X ≥ 1) > 0.90

Compute the value of n as follows:

$P (X \geq 1) > 0.90\\\\1-(\frac{5}{6})^{n}>0.90\\\\(\frac{5}{6})^{n}<0.10\\\\n\cdot \ln (\frac{5}{6})<\ln (0.10)\\\\n<\frac{\ln (0.10)}{\ln (5/6)}\\\\n<12.63\\\\n\approx 12$

Thus, the value of n is 12.

(b)

It is provided that the bag contains 5 green balls and 3 yellow balls.

Ronnie and Julie play a game in which they take turns to draw a ball from the bag at random without replacement.

The winner of the game is the first person to draw a yellow ball.

Also provided that Julie draws the first ball.

P (Ronnie Wins) = P (The 1st yellow ball is selected at an even draw)

                          = P (The 1st yellow ball is drawn at 2nd, 4th and 6th draw)

                          = P (1st yellow ball is drawn at 2nd)

                                  + P (1st yellow ball is drawn at 4th)

                                        + P (1st yellow ball is drawn at 6th)

                          $=[\frac{5}{8}\times \frac{3}{7}]+[\frac{5}{8}\times \frac{3}{7}\times \frac{4}{6}\times \frac{2}{5}]+[\frac{5}{8}\times \frac{3}{7}\times \frac{4}{6}\times \frac{2}{5}\times \frac{1}{4}\times 1]\\\\=0.2679+0.0714+0.0179\\\\=0.3572$

Thus, the probability that Ronnie wins the game is 0.3572.