*X* = the number of freshmen selected from the study until one replied "yes" that same-sex couples should have the right to legal marital status.

*X* = the number of patients calling in claiming to have the flu, who actually have the flu.

*X* = 0, 1, 2, ...25

*X*= number of questions answered correctly*X*~*B*$\left(\text{32,}\frac{\text{1}}{\text{3}}\right)$- We are interested in MORE THAN 75% of 32 questions correct. 75% of 32 is 24. We want to find
*P*(*x*> 24). The event "more than 24" is the complement of "less than or equal to 24." *P*(*x*> 24) = 0- The probability of getting more than 75% of the 32 questions correct when randomly guessing is very small and practically zero.

*X*= the number of college and universities that offer online offerings.- 0, 1, 2, …, 13
*X*~*B*(13, 0.96)- 12.48
- 0.0135
*P*(*x*= 12) = 0.3186*P*(*x*= 13) = 0.5882 More likely to get 13.

*X*= the number of fencers who do**not**use the foil as their main weapon- 0, 1, 2, 3,... 25
*X*~*B*(25,0.40)- 10
- 0.0442
- The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising.

*X*= the number of matches- 0, 1, 2, 3
- In dollars: −1, 1, 2, 3
- \frac{1}{2}
- The answer is −0.0787. You lose about eight cents, on average, per game.
- The house has the advantage.

*X*~*B*(15, 0.281)-
- Mean =
*μ*=*np*= 15(0.281) = 4.215 - Standard Deviation =
*σ*= $\sqrt{npq}$ = $\sqrt{15(0.281)(0.719)}$ = 1.7409

- Mean =
*P*(*x*> 5)=1 – 0.7754 = 0.2246*P*(*x*= 3) = 0.1927*P*(*x*= 4) = 0.2259

It is more likely that four people are literate that three people are.

*X*= the number of adults in America who are surveyed until one says he or she will watch the Super Bowl.*X*~*G*(0.40)- 2.5
- 0.0187
- 0.2304

*X*= the number of pages that advertise footwear*X*takes on the values 0, 1, 2, ..., 20*X*~*B*(20, $\frac{29}{192}$)- 3.02
- No
- 0.9997
*X*= the number of pages we must survey until we find one that advertises footwear.*X*~*G*($\frac{29}{192}$)- 0.3881
- 6.6207 pages

*X*~*G*(0.25)-
- Mean =
*μ*= $\frac{1}{p}$ = $\frac{1}{0.25}$ = 4 - Standard Deviation = σ = $\sqrt{\frac{1-p}{{p}^{2}}}$ = $\sqrt{\frac{1-\text{0}\text{.25}}{{0.25}^{2}}}$ ≈ 3.4641

- Mean =
*P*(*x*= 10) = 0.0188*P*(*x*= 20) = 0.0011*P*(*x*≤ 5) = 0.7627

*X*~*P*(5.5);*μ*= 5.5; $\sigma \text{=}\sqrt{5.5}$ ≈ 2.3452*P*(*x*≤ 6) ≈ 0.6860- There is a 15.7% probability that the law staff will receive more calls than they can handle.
*P*(*x*> 8) = 1 –*P*(*x*≤ 8) ≈ 1 – 0.8944 = 0.1056

Let *X* = the number of defective bulbs in a string.

Using the Poisson distribution:

*μ*=*np*= 100(0.03) = 3*X*~*P*(3)*P*(*x*≤ 4) ≈ 0.8153

Using the binomial distribution:

*X*~*B*(100, 0.03)*P*(*x*≤ 4) = 0.8179

The Poisson approximation is very good—the difference between the probabilities is only 0.0026.

*X*= the number of fortune cookies that have an extra fortune- 0, 1, 2, 3,... 144
- 4.32
- 0.0124 or 0.0133
- 0.6300 or 0.6264
- As
*n*gets larger, the probabilities get closer together.