Respuesta :
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Timothy creates a game in which the player rolls 4 dice.
This means that [tex]n = 4[/tex]
The dice can land in 6 numbers, one of which is 5.
This means that [tex]p = \frac{1}{6}[/tex]
What is the probability in this game of having exactly two dice or more land on a five?
[tex]P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{4,2}.(\frac{1}{6})^{2}.(\frac{5}{6})^{2} = 0.116[/tex]
[tex]P(X = 3) = C_{4,2}.(\frac{1}{6})^{3}.(\frac{5}{6})^{1} = 0.015[/tex]
[tex]P(X = 4) = C_{4,4}.(\frac{1}{6})^{4}.(\frac{5}{6})^{0} = 0.001[/tex]
[tex]P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.116 + 0.015 + 0.001 = 0.132[/tex]
So the correct answer is:
B. 0.132
