Posted by **Samir** on Sunday, January 27, 2008 at 12:14pm.

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?

- Data managment math (#1 of 7) -
**drwls**, Sunday, January 27, 2008 at 12:37pm
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

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