Two cards are drawn at random with replacement from a deck of 52 cards. What is the probability that both cards are aces?
2007-09-16 14:37:16 · 8 answers · asked by vanilla_latte62 1 in Science & Mathematics ➔ Mathematics
P (1st card is an ace) = 4 / 52 = 1 / 13
P (2nd card is an ace) = 4 / 52 = 1 / 13
P (both aces) = (1/13) (1/13) = 1 / 169
2007-09-20 01:21:41 · answer #1 · answered by Como 7 · 0⤊ 0⤋
If this is a truly random drawing, and I assume they are drawn one at a time, then the probability is defined by the likelihood of getting an ace each of the two times.
Since there are 4 aces in a 52 card deck, the probability of drawing one is 4/52, or 1/13. Drawing two results in the probability of 4/52 * 4/52, or 1/(13*13) = 1/169
2007-09-16 14:43:00 · answer #2 · answered by Andrew M 2 · 1⤊ 0⤋
Your chance of picking an ace out of a standard pack of cards is 4/52, if you did get an ace and you did not replace it your chance of picking another ace is 3/51. Now multiply the two probabilities into each other: 4/52 x 3/51 Remember when multiplying fractions you can simplify either vertically or diagonally, NEVER horizontally, so simplify the 4 and 52: 1/13 x 3/51 Now simplify the 3 and 51: 1/13 x 1/17 = 1/221
2016-05-21 05:40:56 · answer #3 · answered by ? 3 · 0⤊ 0⤋
The probability of drawing an ace is 4/52 = 1/13.
The probability of drawing two aces = 1/13 * 1/13 = 1/169.
2007-09-16 14:42:26 · answer #4 · answered by cjcourt 4 · 0⤊ 0⤋
The probability of both is 1:169.
There are 4 aces in a deck. To calculate this, multiply 4/52 by 4/52 and you get 16:2704. If you reduce this, you get 1:169.
2007-09-16 14:45:39 · answer #5 · answered by Scott S 3 · 0⤊ 0⤋
Let X be the number of Ace's drawn. X has the binomial distribution with n = 2 trials and success probability of 1/13.
In general, if X has the binomial distribution with n trials and a success probability of p then
P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, ..., n
P[X = x] = 0 for any other value of x.
for this question you have to find P(X = 2)
plug in the values into the probability mass function above and you'll find:
P(X = 2) = 0.00591716
2007-09-16 16:43:52 · answer #6 · answered by Merlyn 7 · 0⤊ 0⤋
To draw one ace is 1/13. This is true both times because there is replacement.
For both events to happen you multiply the prob so
1/13 X1/13 =1/169
2007-09-16 14:43:34 · answer #7 · answered by Cindy B 5 · 0⤊ 0⤋
221 to 1
4/52 * 3/51
2007-09-16 14:44:21 · answer #8 · answered by the_chosen_one 3 · 0⤊ 0⤋
1/169
(1/13)^2
2007-09-16 14:43:43 · answer #9 · answered by Anonymous · 0⤊ 0⤋