Question

An individual can win a bouncy ball by guessing under which one of four cups the ball is located. After each guess, if the ball is won, a new ball is placed randomly under one of the four cups. If the ball is not won, then the ball is again placed randomly under one of the four cups. If an individual makes five guesses, what is the probability the individual will win a prize exactly two times?

Answer 1

Answer:

I think that it is: [5 2] (.25)^2(.75)^3

Step-by-step explanation:

This looks like a binomial distribution problem. The formula for that is:

Answer 2

A common discrete distribution is used in statistics. The probability the individual will win a prize exactly two times out of five is 0.2637.

What is Binomial distribution?

A common discrete distribution is used in statistics, as opposed to a continuous distribution is called a Binomial distribution. It is given by the formula,

$P(x) = ^nC_x p^xq^{(n-x)}$

Where,

x is the number of successes needed,

n is the number of trials or sample size,

p is the probability of a single success, and

q is the probability of a single failure.

The probability of getting a ball out of the four cups will be,

$P = \dfrac14$

The probability of not getting the ball out of the four cups will be,

$q = \dfrac34$

Now, using the binomial distribution the probability the individual will win a prize exactly two times out of five can be written as,

$P(x) = ^nC_x p^xq^{(n-x)}\\\\P(x=2) = ^5C_2 \cdot (\dfrac{1}{4})^2\cdot (\dfrac34)^{(5-2)}\\\\P(x=2) = 10 \cdot \dfrac{1}{16} \cdot \dfrac{27}{64}\\\\P(x=2)=0.2637$

Hence, the probability the individual will win a prize exactly two times out of five is 0.2637.

Learn more about Binomial Distribution:

https://brainly.com/question/14565246

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