Question

Timothy creates a game in which the player rolls 4 dice. What is the probability in this game of having exactly two dice or more land on a five?
A. 0.016
B. 0.132
C. 0.868
D. 0.984

Answer 1

Answer:

B. 0.132

Step-by-step explanation:

For each time the dice is thrown, there are only two possible outcomes. Either it lands on a five, or it does not. The probability of a throw landing on a five is independent of other throws. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Timothy creates a game in which the player rolls 4 dice.

This means that $n = 4$

The dice can land in 6 numbers, one of which is 5.

This means that $p = \frac{1}{6}$

What is the probability in this game of having exactly two dice or more land on a five?

$P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{4,2}.(\frac{1}{6})^{2}.(\frac{5}{6})^{2} = 0.116$

$P(X = 3) = C_{4,2}.(\frac{1}{6})^{3}.(\frac{5}{6})^{1} = 0.015$

$P(X = 4) = C_{4,4}.(\frac{1}{6})^{4}.(\frac{5}{6})^{0} = 0.001$

$P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.116 + 0.015 + 0.001 = 0.132$

So the correct answer is:

B. 0.132