1: What is the probability that at least two of the 12 councillors elected in the next municipal election will have the same birthdays on the same day?
2:Hoyda has 15 percent of the car market. If a random sample of 20 auto mobiles is conducted what is the probabilitity that (a) exactly one of the car was made by hoyda (b) atleast two of the car were made by hoyda.
3:A basketball game in a league between the sharks and the jets was scheduled for thursday night. Only four players from the sharks showed up and so the jets, who had six players won by default, so the evening would not be spoiled ,the players arrieved four the game decided to have a pickup scrimmage and picked the two teams by drawing names from a hat,what is the probability that rhe four sharks end up on the same team?
4: the odds againsts winning in craps are 507 to 493. what is the expected return on a 20.00 wager?
5: In the best-of -five hockey series , the probability of winning the next game increases for each team by 0.1 if the previous game was won. If the teams ,the Ironmen and the Mustangs ,are in such a series and at the start of the series are evenly matched, what is the probability that (a) the ironmen will win in three straight? (b) the Mustang win the series?
6: Find the number of ways in which at least one piece of fruit could be chosen from a basket containing 4 apples, 5 banana ,2 canta loupes, and 3 pears?
7: The basket ball team has total 14 players : 3-1st year player,5-2nd year player and 6-3rd year player (a) in how many ways can the coach choose a starting line up(5 players) with at least one 1 st player. (b) in how many ways can he set up a starting lineup with two 2nd year and three 3rd year player?
Please post one question at a time and show your work. This is not a place where homework is done for you. There are other sites where that is done for a fee, but you won't learn much that way.
1. The probability that the first person in a group of 1 has no match is 365/365. The probability that the second person is not a match to the first is 364/365. The probability that the third is not a match to either of the first two is 363/365. Continue this logic and the probability that there are no matches among 12 people is
1*(364/365)*(363/365)*(362/365)*(361/365)*(360/365)*(359/365)*(358/365)*(357/365)*(356/365)*(355/365)*(354/365)= 0.833 (check my math)
1-0.833 = 0.167 is the probability that there is at least one pair with the same birthday
1: To find the probability that at least two of the 12 councillors elected in the next municipal election will have the same birthdays on the same day, we can use the concept of the birthday paradox.
The probability that none of the councillors share the same birthday is calculated as follows:
P(no shared birthdays) = (365/365) * (364/365) * (363/365) * ... * (355/365)
This is because the first councillor can have any birthday, the second councillor should have a different birthday (out of the 364 remaining days), the third should have a different birthday (out of the 363 remaining days), and so on.
Therefore, the probability that at least two councillors share the same birthday is:
P(at least two shared birthdays) = 1 - P(no shared birthdays)
We can calculate this value using the formula above.
2: To calculate the probability that exactly one of the 20 automobiles in a random sample was made by Hoyda, we need to use the binomial probability formula.
The formula for the probability of exactly x successes in n trials, where the probability of success in each trial is p, is given by:
P(x) = (nCx) * (p^x) * ((1-p)^(n-x))
In this case, x=1, n=20, and p=0.15 (probability of a car being made by Hoyda).
To calculate the probability that at least two of the cars in the sample were made by Hoyda, we need to calculate the probabilities for x=2, x=3, and so on, up to x=20, and sum them all.
3: To calculate the probability that the four sharks end up on the same team after a pickup scrimmage, we can use the concept of combinations.
There are 10 players in total (4 sharks and 6 jets), and we need to select a team of 4 players. The total number of possible teams is the combination of 10 players taken 4 at a time (10C4).
The probability that the four sharks end up on the same team is the number of favorable outcomes (where all four sharks are selected) divided by the total number of possible outcomes.
4: To find the expected return on a $20.00 wager in craps, we need to calculate the expected value. The expected value is the probability-weighted sum of all possible outcomes.
In this case, the probability of winning is 493/(493+507) and the probability of losing is 507/(493+507). The amount won if we win is $20.00 and the amount lost if we lose is -$20.00.
To calculate the expected return, we multiply the probability of each outcome by its corresponding amount and sum them all.
5: To calculate the probability that the Ironmen will win in three straight games in a best-of-five hockey series, we need to consider the probabilities of winning each individual game.
If the Ironmen win the first three games, the probability is given by the product of the probabilities of winning each game. If the Mustangs win the series, they can do so by winning either 3 games, 4 games, or 5 games. We need to calculate the probabilities for each of these outcomes and sum them.
6: To find the number of ways in which at least one piece of fruit could be chosen from a basket containing 4 apples, 5 bananas, 2 cantaloupes, and 3 pears, we can use the principle of inclusion-exclusion.
First, find the total number of ways to choose fruits from the basket, which is the sum of choosing from each type of fruit separately. Then, subtract the cases where no fruit is chosen from any type. Finally, add back the cases where there is overlap between fruit types (for example, choosing both apples and bananas).
7: To calculate the number of ways the coach can choose a starting lineup of 5 players from a basketball team with 3 first-year players, 5 second-year players, and 6 third-year players, we can use combinations.
For (a), we need to make sure at least one first-year player is chosen while selecting the remaining 4 players from the rest. We can calculate this by finding the number of combinations of choosing 1 first-year player and 4 players from the remaining 18 (3C1 * 18C4).
For (b), we need to select 2 second-year players and 3 third-year players. We can calculate this by finding the number of combinations of choosing 2 second-year players and 3 third-year players (5C2 * 6C3).