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.
You have a total of 11 letters in "mathematics" and you have 4 vowels (a,e,a,i) so the probability of drawing a vowel is 4/11. In other words if you were to consider the vowel's to be 1's and the consonants 2's. What is the probability of drawing a "1". There would be 4 1's and 7 2's. It would be 4/11
The probability of drawing 3 hearts in a row in a 52 card deck is 13/52 * 12/51 * 11/50 = 1716/132600 = 0.01294, or approximately 1 chance in 77.
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.
There are 11 numbers from 10 to 20 if we include 10 and 20. Of those, only 16 is a perfect square. So the probability if 1/11 or .09 repeating.
The probability that the sum of the numbers rolled is either even or a multiple of 5 is 11/18.
The probability of drawing the first face card is 12 in 52. The probability of drawing the second is 11 in 51. The probability of drawing the third is 10 in 50. Thus, the probability of drawing three face cards is (12 in 52) times (11 in 51) times (10 in 50) or (1320 in 132600) or about 0.009955.
You have a total of 11 letters in "mathematics" and you have 4 vowels (a,e,a,i) so the probability of drawing a vowel is 4/11. In other words if you were to consider the vowel's to be 1's and the consonants 2's. What is the probability of drawing a "1". There would be 4 1's and 7 2's. It would be 4/11
The probability is 11/21.
The probability is 22/52 = 11/26.
The probability of drawing a spade is 1/4 (13/52). The probability of drawing a face card is 3/13 (12/52). The probability of drawing a spade face card is 3/52 (3/52). 13 + 12 - 3 = 22 The probability of drawing a spade or a face card is 11/26 (22/52).
To find the probability of drawing a red marble first and then a blue marble, we first calculate the probability of each event separately. The probability of drawing a red marble is ( \frac{3}{11} ), since there are 3 red marbles out of a total of 11 marbles. After returning the red marble, the probability of then drawing a blue marble is ( \frac{5}{11} ). Therefore, the combined probability of drawing a red marble first and then a blue marble is ( \frac{3}{11} \times \frac{5}{11} = \frac{15}{121} ).
The probability is 11/13.
The probability of drawing 3 hearts in a row in a 52 card deck is 13/52 * 12/51 * 11/50 = 1716/132600 = 0.01294, or approximately 1 chance in 77.
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.
The probability of drawing a red even card is 5 in 13 (assuming that the face cards are neither even nor odd). The probability of throwing a sum of 11 is 1 in 18. The probability, then, of doing both of these actions, since they are mutually independent, is simply their product, or (5 in 13) times (1 in 18) equals 5 in 234, or about 0.02137.
If the probability of an event will occur is p, then the probability that it will occur in n trials is pn.(That's p raised to the n power). So if you bet on 12 numbers, then (38-12=26) numbers are empty. The probability of the ball landing on one of these empty numbers is 26/38. So (26/38)^11 = 0.01538, which is about 1.538 % or a 1 in 65 chance.
1 in 12 * 11 / 2 / 1 or 1 in 66.