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Can anyone help me find each probalilities of each events?

 

Event 4: We draw five cards with replacement. We get five aces. (1/52**5)

 

Event 5: We draw five cards without replacement. The five cards are from the same suit. (1/508)

 

Event 6: We draw five cards without replacement. The five cards make a sequence of cards of consecutive order. Example: 10, jack, 9, 8, 7. 

 

Event 7: We draw five cards without replacement. The five cards make a full house: three cards of one rank and two cards of different rank. 

 

Thanks.(I posted these problems two hours ago,but no one anwser.sadI really need these answerfor my classes.)frown

 Nov 17, 2016
 #1
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Event 4: We draw five cards with replacement. We get five aces.

 

The probability of getting an ace on each draw remains the same  = (4/52) = (1/13)

 

So....the probability  of doing this 5 times in a row =  (1/13)^5  = 1/ 371,293

 

 

Event 5: We draw five cards without replacement. The five cards are from the same suit. (1/508)

 

This is known as a "flush"....we have four ways to do this and from any of the suites, we want to choose 5 of 13 cards....so we have   C(4,1) * C (13,5)  = 5148 ways of doing this...and the number of ways of choosing any 5 cards from 52 = C(52,5) = 2,598,560....so....the probability = 5148 / 2,598,560 = about 0.00198

 

 

Event 6: We draw five cards without replacement. The five cards make a sequence of cards of consecutive order. Example: 10, jack, 9, 8, 7. 

 

This is known as a "straight"  .....  There are 10 possible "straights" [ if we can make the ace either high or low] ....and from each rank, we want to choose 1 card out of 4.......and we want to do this 5 times ...so....the total number of ways to do this = 10 * C(4,1)^5 =  10,240 .......so the probability = 10,240 / 2,598,560 = about .00394  

 

 

Event 7: We draw five cards without replacement. The five cards make a full house: three cards of one rank and two cards of different rank. 

 

We want to coose 3 of 4 cards out of 13 ranks and 2 of 4 cards out out the remaing 12 ranks...so we have

 

13* C(4,3) * 12 * C(4,2) = 3744 possibe ways.....so...the probability = 3744/ 2,598,560 = about .00144

 

 

 

cool cool cool

 Nov 17, 2016

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