2.10 Binomial Counts: Model, Calculate and Interpret — Answers and Rubrics

X counts defects in fixed n=20 independent trials. Each component is defect (success) or not defect (failure), with constant p=.02 and 1−p=.98 under the stated stable model. Hence X is binomial with n=20,p=.02.

Y has a fixed sample size and binary outcomes, but drawing without replacement from this small box changes the defect probability and makes draws dependent: after a defective first draw the next defect probability is 11/39; after a nondefective first draw it is 12/39. Z counts trials until the third success, so its trial count is not fixed in advance and it is not a success count over a fixed n.

$\mu_X=np=25(.20)=5$ redemptions. $V(X)=np(1-p)=25(.20)(.80)=4$ squared redemptions, and $\sigma_X=\sqrt4=2$ redemptions.

Across many 25-coupon campaigns, the average redemption count approaches five per campaign. Campaign counts typically differ about two redemptions from that mean. About 10.91228% of such campaigns have at least eight redemptions. Five is an expected count for a set of 25; individual campaigns vary and a single coupon’s probability remains .20.

The empirical probabilities for X=0,1,2,3,4 are .27,.45,.21,.07,0. They total 1. The estimate of P(X≥2) is (.21+.07+0)=28/100=.28: 28% of these simulated four-play groups had at least two prizes.

Valid runs differ. Each recorded trial must contain four independent outcomes with prize probability .25, for example four uniform digits 0–3 with prize on 0. Count every X value and divide by 50; add counts for X=2,3,4 and divide by 50 for the event estimate. The table must total 50 and probabilities must total 1. Finite-run variation can change the estimate while the model stays fixed.

$P(X=x)=\binom4x(.25)^x(.75)^{4-x}$. For x=0,1,2,3,4, the probabilities are .31640625,.421875,.2109375,.046875,.00390625. They total 1. The exact P(X≥2)=.26171875, while .28 is the supplied finite simulation estimate; the difference is about .01828.

$P(X=x)=\binom8x(.30)^x(.70)^{8-x}$ for x=0,1,2,…,8. $P(X=3)=56(.30)^3(.70)^5=.25412184$. About 25.41% of repeated eight-visit routes have exactly three visits needing extra parts.

$P(X\ge3)=\sum_{x=3}^{8}\binom8x(.30)^x(.70)^{8-x}=1-P(X\le2)=.44822619$. $P(X>3)=\sum_{x=4}^{8}\binom8x(.30)^x(.70)^{8-x}=1-P(X\le3)=.19410435$. The first includes x=3 and the second excludes it; the difference is .25412184.

$P(2\le X\le4)=P(X=2)+P(X=3)+P(X=4)=.29647548+.25412184+.13613670=.68673402$. At the endpoints, $P(X=0)=\binom80(.3)^0(.7)^8=.05764801$, and $P(X=8)=\binom88(.3)^8(.7)^0=.00006561$. Each is in [0,1].