Question

Having trouble with these two problems

3. Suppose a Coed basketball team consists of 7 females and 8 males. Suppose a starting line-up of five players is randomly selected (each possible starting line-up is equally likely). What is the exact probability that the chosen starting line-up contains at least one male? A bar is trying to get a feel for the number of drinks that a person will order. Let the random variable X denote the number of drinks a randomly chosen customer will buy. From past experience the bar knows that 20% ofpeople will order l drink, 30% will order 2 drinks, 30% order 3 drinks, and 10% order 4 drinks. If a customer may buy 0, 1, 2, 3, or 4 drinks, then answer the following questions (a) Use a table to represent the PDF for the random variable X (b) Find P(X<4); that is, the exact probability that the random variable X takes on a value less than 4. (c) Find the exact expected value of X. d) Using English sentence(s) explain the meaning of the numerical value you got in part (c). (e) Find the standard deviation of X (round your answer to four decimal places)

Answer 1

3. Team Consists of 7 females and 8 males.

Total number of players = 15

Starting lineup : 5 players

Probability that the chosen starting lineup contains at least one male = 1 - Probability that the chosen starting lineup contains no male(all females)

Number of ways choosing 5 players for starting lineup from 15 players available = = 3003

15

For a all female lineup : Number of ways choosing 5 players for starting lineup from 7 female players = =21

75

Probability that the chosen starting lineup contains no male(all females) = 21/3003

Probability that the chosen starting lineup contains at least one male = 1 - Probability that the chosen starting lineup contains no male(all females) = 1 -21/3003 = 2982/3003

Probability that the chosen starting lineup contains at least one male = 2982/3003

4.Given

X: Number of drinks	Past Experience: Percent of people
0
1	20%
2	30%
3	30%
4	10%

Percent of people order '0' drinks = 100 - (20+30+30+10) = 10

Converting the % in probabilities we get the following:

PDF of random variable X

X: Number of drinks	P(x)
0	0.1
1	0.2
2	0.3
3	0.3
4	0.1

P(X<4) = P(X=0)+P(X=1)+P(X=2)+P(X=3) = 0.1+0.2+0.3+0.3 = 0.9

Expected Value of X : E(X)

$\small E(X) = \sum xP(x)$

0 x 0.1 1 x 0.2 + 2 x 0.3 + 3 x 0.3 + 4 x 0.1 0+0.2 + 0.6 + 0.9 + 0.4 = 2.1

Exact expected value of X = 2.1

(d)

On an average each customer will drink 2.1 drinks

(e)

Standard deviation of X :

$\small \sqrt{E(X^{2})-E(X)^{2}}$

E(X2)

$\small = 0^{2}\times 0.1 + 1^{2}\times 0.2 + 2^{2}\times 0.3 + 3^{2}\times 0.3+4^{2}\times 0.1 = 0+0.2+1.2+2.7+1.6=5.7$

Standard deviation of X :

VE (X2)一E(x)-= v5.7-2.12 = V1.29 = 1.1358

Standard deviation of X = 1.1358