For 9 to 11 odds of winning;
Probability of winning:
0.55, or;
Chance of winning:
55%
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The probability of drawing one face card is 12 in 52. The probability of drawing the second face card, then, is 11 in 51. And so on, 10 in 50, 9 in 49, and 8 in 48. Since this is sequential probability, simply multiply the odds, giving (12/52)(11/51)(10/50)(9/49)(8/48) or (95040/311875200). I'l leave it to the reader to do the simple reduction to lowest terms.
What the answer
In my family,my dad,myself and my brother all share the same birthday.(9/11/24; 9/11/52; and 9/11/53). Also,my niece was born on 9/11/85.Three generations! Can anyone beat that?
If the number 8 is PUT BACK INTO the group after the first draw, the probability of drawing an 8 then a 2 from 11 numbers is: 0.8% or eight-tenths of a percent [reason #1]If the number 8 is REMOVED from the group after the first draw, the probability of drawing an 8 then a 2 from 11 numbers is: 0.9% or nine-tenths of a percent [reason #2]Reason # 1 :On the first draw, you have a 1 in 11 chance of drawing the 8 (9%) then on the second draw there is again a 1 in 11 chance of drawing the 2 (9%). So 9% x 9% = 0.8% or 8/10 % (or 0.008)Reason # 2 :On the first draw, you have a 1 in 11 chance of drawing the 8 (9%) then on the second draw you only have a 1 in 10 chance of drawing the 2 (10%). So 9% x 10% = 0.9% or 9/10 % (or 0.009) OR If the first ball is removed, the odds are 55 to 1.
First we find the probability of getting a 7. Of the 36 outcomes possible 6 result in a sum of 7, in other words 1/6. The probability of getting an 11 is 2/36 or 1/18. The probability of getting one or the other is the sum of the two, 8/36 or 2/9. The proability of getting neither is equal to the probability of getting anything other than 7 or 8. We find this value by subtracting 2/9 from 1. So the probability of not getting 7 or 11 is 7/9.