Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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Chapter 5.4, Problem 2E
To determine

(a)

To find:

The specified probability for the given scenario.

Expert Solution
Check Mark

Answer to Problem 2E

Solution:

Value of the required probability is 0.2742.

Explanation of Solution

Formula used:

For a hyper geometric random variable X, the probability of obtaining r successes in n dependent trials is given by,

P(X=r)=(Ckr)(CNknr)(CNn)

Here, N is the number of items in the entire population,

n, is the number of trials,

k, is the number of successes in the entire population, and

r, is the number of success obtained in n trials.

The formula to calculate Ckr is,

Ckr=k!r!(kr)!

Calculation:

Total number of gift is 10.

N=10

Consider the event “gift is for Abby” as a success.

There are 5 gifts for Abby.

k=5

Number of trial is 4.

n=4.

Compute the probability for exactly 1 success.

Substitute 1 for r, 5 for k, 4 for n, and 10 for N in the formula for exactly r successes.

P(X=1)=(C51)(C10541)(C104)=(C51)(C53)(C104)=5!1!4!×5!3!2!10!4!6!=5×10210

Further simplify the value of P(X=1).

P(X=1)=5×10210=0.2380952381

P(X=1)=0.2380

Conclusion:

Thus, the probability of getting one gift for Abby is 0.2380.

To determine

(b)

To find:

The specified probability for the given scenario.

Expert Solution
Check Mark

Answer to Problem 2E

Solution:

Value of the required probability is 0.8968.

Explanation of Solution

Formula used:

For a hyper geometric random variable X, the probability of obtaining at most r successes in n dependent trials is given by,

P(Xr)=P(X=0)+P(X=1)+P(X=2)+...+P(X=r)

P(X=r)=(Ckr)(CNknr)(CNn)

Here, N is the number of items in the entire population,

n, is the number of trials,

k, is the number of successes in the entire population, and

r, is the maximum number of successes obtained in n trials.

The formula to calculate Ckr is,

Ckr=k!r!(kr)!

Calculation:

Total number of gift is 10.

N=10

Consider the event “gift is for Andrew” as a success.

There are 5 gifts for Andrew.

k=5

Number of trial is 5.

n=5.

Compute the probability for at most 3 successes.

Substitute 3 for r in the formula for at most r successes.

P(X3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)

Substitute 0 for r, 5 for k, 5 for n, and 10 for N in the formula for P(X=r) to calculate P(X=0).

P(X=0)=(C50)(C10550)(C105)=(C50)(C55)(C105)=5!0!5!×5!5!5!10!5!5!=1×1252

Further simplify the value of P(X=0).

P(X=0)=1×12520.003968

P(X=0)=0.003968

Substitute 1 for r, 5 for k, 5 for n, and 10 for N in the formula for P(X=r) to calculate P(X=1).

P(X=1)=(C51)(C10551)(C105)=(C51)(C54)(C105)=5!1!4!×5!4!1!10!5!5!=5×5252

Further simplify the value of P(X=1).

P(X=1)=252520.099206

P(X=1)=0.099206

Substitute 2 for r, 5 for k, 5 for n, and 10 for N in the formula for P(X=r) to calculate P(X=2).

P(X=2)=(C52)(C10552)(C105)=(C52)(C53)(C105)=5!2!3!×5!3!2!10!5!5!=10×10252

Further simplify the value of P(X=2).

P(X=2)=1002520.3968254

P(X=2)=0.3968254

Substitute 3 for r, 5 for k, 5 for n, and 10 for N in the formula for P(X=r) to calculate P(X=3).

P(X=3)=(C53)(C10553)(C105)=(C53)(C52)(C105)=5!3!2!×5!2!3!10!5!5!=10×10252

Further simplify the value of P(X=3).

P(X=3)=1002520.3968254

P(X=3)=0.3968254

Add the values of P(X=0),P(X=1),P(X=2),andP(X=3) to get the value of P(X3).

P(X3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=0.003968+0.099206+0.3968254+0.39682540.8968

P(X3)=0.8968

Conclusion:

Thus, the probability of getting at most 3 gifts for Andrew is 0.8968.

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Chapter 5 Solutions

Beginning Statistics, 2nd Edition

Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.1 - Prob. 17ECh. 5.1 - Prob. 18ECh. 5.1 - Prob. 19ECh. 5.1 - Prob. 20ECh. 5.1 - Prob. 21ECh. 5.1 - Prob. 22ECh. 5.1 - Prob. 23ECh. 5.2 - Prob. 1ECh. 5.2 - Prob. 2ECh. 5.2 - Prob. 3ECh. 5.2 - Prob. 4ECh. 5.2 - Prob. 5ECh. 5.2 - Prob. 6ECh. 5.2 - Prob. 7ECh. 5.2 - Prob. 8ECh. 5.2 - Prob. 9ECh. 5.2 - Prob. 10ECh. 5.2 - Prob. 11ECh. 5.2 - Prob. 12ECh. 5.2 - Prob. 13ECh. 5.2 - Prob. 14ECh. 5.2 - Prob. 15ECh. 5.2 - Prob. 16ECh. 5.2 - Prob. 17ECh. 5.2 - Prob. 18ECh. 5.2 - Prob. 19ECh. 5.2 - Prob. 20ECh. 5.2 - Prob. 21ECh. 5.2 - Prob. 22ECh. 5.2 - Prob. 23ECh. 5.2 - Prob. 24ECh. 5.2 - Prob. 25ECh. 5.3 - Prob. 1ECh. 5.3 - Prob. 2ECh. 5.3 - Prob. 3ECh. 5.3 - Prob. 4ECh. 5.3 - Prob. 5ECh. 5.3 - Prob. 6ECh. 5.3 - Prob. 7ECh. 5.3 - Prob. 8ECh. 5.3 - Prob. 9ECh. 5.3 - Prob. 10ECh. 5.3 - Prob. 11ECh. 5.3 - Prob. 12ECh. 5.3 - Prob. 13ECh. 5.3 - Prob. 14ECh. 5.3 - Prob. 15ECh. 5.3 - Prob. 16ECh. 5.3 - Prob. 17ECh. 5.3 - Prob. 18ECh. 5.3 - Prob. 19ECh. 5.3 - Prob. 20ECh. 5.3 - Prob. 21ECh. 5.3 - Prob. 22ECh. 5.3 - Prob. 23ECh. 5.3 - Prob. 24ECh. 5.4 - Prob. 1ECh. 5.4 - Prob. 2ECh. 5.4 - Prob. 3ECh. 5.4 - Prob. 4ECh. 5.4 - Prob. 5ECh. 5.4 - Prob. 6ECh. 5.4 - Prob. 7ECh. 5.4 - Prob. 8ECh. 5.4 - Prob. 9ECh. 5.4 - Prob. 10ECh. 5.4 - Prob. 11ECh. 5.4 - Prob. 12ECh. 5.4 - Prob. 13ECh. 5.4 - Prob. 14ECh. 5.4 - Prob. 15ECh. 5.4 - Prob. 16ECh. 5.4 - Prob. 17ECh. 5.4 - Prob. 18ECh. 5.4 - Prob. 19ECh. 5.CR - Prob. 1CRCh. 5.CR - Prob. 2CRCh. 5.CR - Prob. 3CRCh. 5.CR - Prob. 4CRCh. 5.CR - Prob. 5CRCh. 5.CR - Prob. 6CRCh. 5.CR - Prob. 7CRCh. 5.CR - Prob. 8CRCh. 5.CR - Prob. 9CRCh. 5.CR - Prob. 10CRCh. 5.CR - Prob. 11CRCh. 5.CR - Prob. 12CRCh. 5.CR - Prob. 13CRCh. 5.P - Prob. 1PCh. 5.P - Prob. 2PCh. 5.P - Prob. 3PCh. 5.P - Prob. 4PCh. 5.P - Prob. 5PCh. 5.P - Prob. 6PCh. 5.P - Prob. 7PCh. 5.P - Prob. 8PCh. 5.P - Prob. 9PCh. 5.P - Prob. 10PCh. 5.P - Prob. 11PCh. 5.P - Prob. 12PCh. 5.P - Prob. 13PCh. 5.P - Prob. 14PCh. 5.P - Prob. 15PCh. 5.P - Prob. 16PCh. 5.P - Prob. 17P
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