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can someone answer these another problems?
1. A pair of dice is tossed. Find the probability of getting a) a total of 8? B) at most a total of 5? 2.Two cards are drawn in succession from a deck without replacement. What is the probability that both cards are greater than 2 and less than 8? 3. If each coded item in catalog begins with 5 different letters and 3 distinct nonzero digits, find the probability of randomly selection one of these coded items with the first letter a vowel and last digit even? 4.In a poker hand consisting of cards, find he probability of holding 3 aces? 5.In a game of poker where a player is dealt 5 cards from a deck of 53 playing cards (52 cards + 1 joker), what is the probability of getting a full house? 6.From a group of 8 mathematicians and 6 physicists, a committee consisting of 5 members is to be formed. What is the probability of forming a committee that includes at least two mathematicians and at least 1 physicist?
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- I'll answer the first question. I'll try to edit and add other answers to other questions if possible in my return. Probabilities of totals: 2 & 12 = 1 each or 2 3 & 11 = 2 each or 4 4 & 10 = 3 each or 6 5 & 9 = 4 each or 8 6 & 8 = 5 each or 10 7 = 6 Sum of probabilities: 2 + 4 + 6 + 8 + 10 + 6 = 36 Answer: probability of 8: 5:31 or 5/36 or 13 8/9% or 13.88 8/9%
- 1a. There are 7 ways to get an 8: 17 26 35 44 53 62 71. So the probability is 7/36. 1b. There are 10 ways to get at most a total of 5: 11, 12, 21, 13, 22, 31, 14, 23, 32, 41, so the probability is 10/36 = 5/18. 2. There are 5 cards in each suit greater than 2, but less than 8, so the probability the first card is there is 5/13. The probability the second card is greater than 2 but less than 8 is (20-1)/52 = 19/52, since there's no replacement. So the probability is (5/13)*(19/52)=95/676. 3. There are 26*25*24*23*22*9*8*7 different codes. There are 5*25*24*23*22*4*8*7 different codes with the first letter a vowel and the last digit even. So the probability is: 5*25*24*23*22*4*8*7/26*25*24*23*22*9*8*7 = 5/26*4/9 = 10/117. 4. I'll assume you meant 5 cards. There are (4 choose 3)*(49 choose 2) hands with 3 aces. There are (52 choose 5) hands all together. So the probability is: (4 choose 3)*(49 choose 2)/ (52 choose 5) = 2/1105. 5. There are (13 choose 2)*2*(4 choose 3)*(4 choose 2)=3744 full houses without a joker (choose 2 ranks, then choose which has 3 cards, and which has 2, then choose 3 out of the 4 possible, and 2 out the 4 possible for each rank). There are (13 choose 2)*(4 choose 2)*(4 choose 2)=2808 full houses with a joker (choose 2 ranks, then choose 2 cards from each rank). For a total of 6552 full houses. There are (53 choose 5) = 2869685 hands. So the probability is 6552/286985 = 72/31535 or about 0.2283% 6. Find how many committees contain less than 2 mathematicians or less than 1 physicist. There are (8 choose 1)*(6 choose 4)=120 committees with 1 mathematician, and (6 choose 5)=6 committees with 0 mathematicians, and there are (8 choose 5)=56 committees with 0 physicists. So there are 182 total committees which don't satisfy the condition. There are (14 choose 5)=2002 committees in total. So the probabilty of containing at least 2 mathematicians and at least 1 physicist is 1 - 182/2002 = 10/11.
