1. ## Statistics URGENT!! HELP!!

Hey guys, I'm just in major need of help trying to pass this hard stats class and I got 7 out of 10 right from my quiz and I need three more right so if any of you are bored and would like to figure them out for me this girl would be much appreciated for your help!! Id love you forever!!

1. An airline estimates that 98% of people booked their flights actually show up. If the airline books 76 people on a flight for which the maximum number is 74, what is the probability that the number of people who show up will exceed the capacity if the plane?

2. The participants in a television quiz show are picked from a large pool of applicants with approximately equal numbers of men and women. Among the last 11 participants there have been only 2 women. If participants are picked randomly, what is the probability of getting 2 or fewer women when 11 people are picked?

3. A car insurance company has determined the 7% of all drivers were involved in a car accident last year. Among the 11 drivers living on one particular street. 3 were involved in a car accident last year. if 11 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year?

Thanks guys!

2. 1. An airline estimates that 98% of people booked their flights actually show up. If the airline books 76 people on a flight for which the maximum number is 74, what is the probability that the number of people who show up will exceed the capacity if the plane?
These appear to be binomial in nature. Binomial probability is very mechanical and patterned.

Out of 76 chosen, what's the probability 75 or 76 show up. Because it must exceed 74.

$\displaystyle \sum_{k=75}^{76}C(76,k)(0.98)^{k}(0.02)^{76-k}$

$\displaystyle C(76,75)(0.98)^{75}(0.02)^{1}+C(76,76)(0.98)^{76}( 0.02)^{0}$

2. The participants in a television quiz show are picked from a large pool of applicants with approximately equal numbers of men and women. Among the last 11 participants there have been only 2 women. If participants are picked randomly, what is the probability of getting 2 or fewer women when 11 people are picked?

The percentage of women is 2/11. Out of 11 chosen, what's the probability of choosing 0, 1, or 2 women.

We have to sum up the prob of choosing 0, 1, or 2 women. Therefore,

$\displaystyle \sum_{k=0}^{2}C(11,k)\left(\frac{2}{11}\right)^{k} \left(\frac{9}{11}\right)^{11-k}$

3. thanks so much for your help!

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### the participants in a television quiz show

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