(1 point) A person is to walk into a casino and play a certain game. The chance
ID: 3057244 • Letter: #
Question
(1 point) A person is to walk into a casino and play a certain game. The chance the person will win the game is 0.38. Once they play the first game, win or lose, they are to play the game 3 more times for a total of 4 games. A random variable X is to count how many of the 4 games the gambler wins (a) Finish the probability distribution of X below. Use four decimals in each of your entries. (b) From the distribution you found in part (a), what can you say about the distribution of X? The distribution of X is? Enter your answers to two decimals.) I. with an mean of ll games won and a standard deviation of games won. Preview My Answers Submit AnswersExplanation / Answer
Back-up Theory
Number of ways of selecting r things out of n things is given by nCr which expands as
(n!)/{(r!)(n – r)!}……………………………………………………………………..(1)
Part (a)
P(X = 0) = P(None of the 4 games is won)
= 0.624[given probability of winning is 0.38, probability of not winning is0.62]
= 0.1478
P(X = 1) = P(Exactly one win)
= P(one win and other three losses)
Now, one win can be any one of the 4 games in 4 ways (4C1). One win probability is 0.38 and three losses is 0.623
Thus, P(X = 1) = (4C1)(0.38)( 0.623)
= 0.3623
Similarly,
P(X = 2) = (4C2)(0.382)( 0.622) = 0.3330
P(X = 3) = (4C3)(0.383)( 0.62) = 0.1361
P(X = 4) = (4C4)(0.384)( 0.620) = 0.0209.
Thus, the probability distribution of X is:
x
0
1
2
3
4
Total
p(x)
0.1478
0.3623
0.3330
0.1361
0.0208
1
ANSWER
Part (b)
The above distribution is
Binomial with mean 1.52 and standard deviation 0.97. ANSWER
DONE
[Additional Note:
If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then
probability mass function (pmf) of X is given by
p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n ………………………………..(1)
[The above probability can also be directly obtained using Excel Function of Binomial Distribution] ……………………………………………………………………………….(1a)
Mean (average) of X = E(X) = np…………………………………………………………..(2)
Variance of X = V(X) = np(1 – p)…………………………………………………………..(3)
Standatd Deviation of X = SD(X) = { np(1 – p)} ………………………………………...(4)
In the given question, n = 4, p = 0.38]
x
0
1
2
3
4
Total
p(x)
0.1478
0.3623
0.3330
0.1361
0.0208
1
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.