course Mth 152
Thanks for all your help. Here are a few more problems that I could use some help with. I'm really anxious about this test on probability.
1) Officer Andrews has been assigned to an area where 25% of drivers speed and can be given a ticket. Find the probability that he will give is first ticket after 7 drivers pass his area.
The first six drivers won't get tickets; probability is .75 each. The 7th will; probability is .25. So the probability that the first six won't and the seventh will is .75^6 * .25.
This is very similar to a situation I addressed as an extra note in my last posting. Take a look and be sure you see the connection.
2) If 80% of scheduled flights actually take place and cancellations are independent events, what is the probability that three separate flights will take place?
Each of the three flights takes place with 80% probability so the probability is .8 * .8 * .8 or .8^3.
3) If 3 boys and 2 girls are to be arranged in a row, what is the probability that 2 boys will not be in adjacent seats?
P(5,5) = 120; 120-4=116; 116/120=0.967 probability
The only way would be for the 3 boys to be in seats 1, 3 and 5.
There are C(5, 3) ways to arrange three boys in 5 seats, and only 1 fulfills the condition, so the probability is 1 / C(5, 3).
4) Numbers is a game where you bet $1.00 on any three-digit number from 000 to 999. If your number comes up, you get $600.00. Find the expected winnings.
600(1/1000) + -600(999/1000)= 600/1000 + -599400/1000 = -598,800/1000 = -598.80 Am I anywhere close on this problem?
Fairly close. However you don't lose $600 if you fail to win; you only lose $1, and even if you don't lose you've paid the dollar.
So you lose $1 every time. 1 time out of 1000 you win $600. The expectation is
$600 * (1/1000) - $1 * 1 = $-0.40.
5) At the first tri-city meeting, there were 8 people from Town A, 7 people from Town B, and 5 people from Town C. If the council consists of 5 people, find the probability that 2 are from Town A, two from Town B, and one from Town C.
C(20,5)=15504
C(8,2)=28; 28/15504 = .002
C(7,2)=21; 21/15504 = .001
C(5,1)=5; 5/15504 = 3.22
.002 + .001 + 3.22 = 3.223 probability
Note that probability can't exceed 1.
To complete the choice you have to choose 2 from Town A and 2 from Town B and 1 from Town C. By the Fundamental Counting Principle you therefore multiply the numbers, you don't add them.
There are C(8, 2) ways to get two from Town A, C(7, 2) ways to get two from Town B and C(5, 1) ways to get 1 from Town C. So there are
C(8, 2) * C(7, 2) * C(5, 1)
ways to choose the 5 people with the specified distribution and since there are C(20, 5) ways to choose the 5 people, the probability is
C(8, 2) * C(7, 2) * C(5, 1) / C(20, 5)
6) It is believed that in a certain town, 35% of the people did not graduate from high school. If we sample adults 18 years or older, what is the probability that the first person who did not graduate from high school will be person number 8?
This is very similar to the first question. The probability is .65^7 * .35.
7) A cookie has less than 8 raisins once out of every 100 cookies. Cookies with less than 8 raisins are rejected. Find the probability that the first cookie rejected will be the fifth cookie?
Again the same reasoning as on #1 gives you probability
.99*4 * .01.
For most students this is the toughest of the tests in either the Mth 151 or 152 course.
`gr9