Question

A fast-food restaurant runs a promotion in which certain food items come with game pieces. According to the restaurant, 1 in 4 game pieces is a winner. If Jeff gets 4 game pieces, what is the probability that he wins exactly 1 prize?
(a) 0.25
(b) 1.00
(c) (41)(0.25)1(0.75)3
(d) (41)(0.25)3(0.75)1
(e) (0.75)3(0.25)1

Answer 1

Answer:

(c) $\left\ ({{4} \atop {1}} )\right.$ $0.25^{1} 0.75^{3}$

Step-by-step explanation:

As given in the statement, we have:

Out of 4 games pieces, 1 is winner.

Probability to win =p= $\frac{1}{4}$

Jeff has game pieces = n = sample size = 4

As we need to find the probability that he wins exactly 1 prize, we will use binomial probability here :

$P (X = k) = \left\ ({{n} \atop {k}} )\right. p^{k} (1-p)^{n-k}$

Evaluating at k=1, (k=1 as we need to find probability for exactly 1 prize won)

put n = 4, p =$\frac{1}{4}$

P (X = 1) =$\left\ ({{4} \atop {1}} )\right. 0.25^{1} (1-0.25)^{4-1}$

P =$\left\ ({{4} \atop {1}} )\right. 0.25^{1} (0.75)^{3}$

Which is the probability that he wins exactly 1 prize and is option c.

Answer 2

Probability (Jeff wins 1 price in 4 game pieces) = C] $(4 c 1)(0.25)^1(0.75)^3$

Important Information : Probability (Winning a price) = 1 / 4 = 0.25

Probability (Not winning price) = 1 - Pr (Winning Price) = 1 - 0.25 = 0.75

Using Binomial Probability : Pr (X = r) = $N c r . P^r . Q^(n-r)$ .

Here N = number of trials (4 game pieces here) , P = Probability of Success (of winning price = 0.25) , R = Number of Success (1 price) , Q = Probability of failure (of not winning price = 0.75) ,

So, Probability = $4 c 1 (0.25)^1 (0.75)^3$

To learn more about Probability, refer https://brainly.com/question/13609688?referrer=searchResults