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A businessman has an important meeting to attend, but he is running a little late. He can take one route to work that has six stoplights or another, longer route that has two stoplights. He figures that if he stops at more than half of the lights on either route, he will be late for the meeting. Assume independence, and assume that the chance of stopping at each light is p =0.5. Which route should he take?

It is claimed that for a particular lottery, 1/10 of the 50 million tickets will win a prize. What is the probability of winning at least one prize if you purchase (a) 10 tickets or (b) 15 tickets?

https://brainmass.com/math/basic-algebra/probability-binomial-distribution-255644

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Solution 4-15

a) Let X denote the number of prizes won if 10 tickets were purchased. Here it is given that the probability of winning a prize is 1/10 = 0.10. Clearly X follows a Binomial distribution with parameter n = 10 and p = 0.10. Thus the p.m.f. of X is given by,
, x = 0, 1, 2, 3,..., 10.
Now, the probability of winning at least one prize is given by,

= 1 - 0.3487
= 0.6513

b) Let X denote the number of prizes won if 15 tickets were purchased. Here it is given that the probability of winning a prize is 0.10. Clearly X follows a Binomial distribution with parameter n = 15 and p = 0.10. Thus the p.m.f. of X is given by,
, x = 0, 1, 2, 3,..., 15.
Now, the probability of winning at least one prize is given by,

= 1 - 0.2059
= 0.7941

Solution 4-16

Let n denote the number of tickets purchased and let X denote the number of prizes won if n tickets were purchased. Here it is given that the probability of winning a prize is 0.10. Clearly X follows a Binomial distribution with parameter n and p = 0.10. Thus the p.m.f. of X is given by,
, x = 0, 1, 2, 3,..., n.
Now, the probability of winning at least one prize is given by,

a) The smallest number of tickets that must be purchased so that the probability of winning at least one prize is greater than 0.50 is the value of n such that
> 0.50
That is, < 1 - 0.50
That is, < 0.50
That is, < log(0.50) [By taking logarithm on both sides]
That is, - 0.10536*n < - 0.69315
That is, 0.10536*n > 0.69315 [The multiplication of -1 reverses the sign of the inequality]
That is, n > 0.69315/0.10536
That is, n > 6.5788
That is, smallest n = 7
Thus the smallest number of tickets that must be purchased so that the probability of winning at least one prize is greater than 0.50 is 7.
b) The smallest number of tickets that must be purchased so that the probability of winning at least one prize is greater than 0.95 is the value of n such that
> 0.95
That is, < 1 - 0.95
That is, < 0.05
That is, < log(0.05) [By taking logarithm on both sides]
That is, - 0.10536*n < - 2.99573
That is, 0.10536*n > 2.99573 [The multiplication of -1 reverses the sign of the inequality]
That is, n > 2.99573/0.10536
That is, n > 28.4332
That is, smallest n = 29
Thus the smallest number of tickets that must be purchased so that the probability of winning at least one prize is greater than 0.95 is 29.

Solution 4-21

Let X denote the number of times the businessman stops at the lights on the route that has 6 stoplights and Y denote the number of times the businessman stops at the lights on the longer route that has 2 stoplights. Here it is given that the probability of stopping at each light is 0.5.
Now, X follows a Binomial distribution with parameter n = 6 and p = 0.5. Thus the p.m.f. of X is given by,
, x = 0, 1, 2, 3,..., 6.
Also, Y follows a Binomial distribution with parameter n = 2 and p = 0.5. Thus the p.m.f. of Y is given by,
, y = 0, 1, 2.
Therefore, the probability that the businessman stops at more than half of the lights on the route that has 6 stoplights is

= 22*0.0156
= 0.3438
Also, the probability that the businessman stops at more than half of the lights on the longer route that has 2 stoplights is

= 0.25

Since the probability that the businessman stops at more than half of the lights on the longer route is minimum, he should take the longer route.

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