I will assume that you mean a five card Poker hand. We can label the cards C1, C2, C3, C4, and C5. We are basically told already that C1 and C2 are both aces. So we have to find the probability that exactly one of C3, C4, and C5 is an ace. Knowing that the first two cards in our hand are both aces means that there are only 50 cards left in the deck. The probability that C3 is an ace and that C4 and C5 are both not aces is (2/50)(48/49)(47/48)=0.03836734694. The same probability also applies to each of C4 and C5, considered independently of each other. Therefore, our final probability is 3* 0.03836734694=0.1151020408
An integer is chosen at random from the first 20,000 positive integers . What is the probability that the integer chosen is divisible by atleast one of the numbers 5,7,8? *
If the balls are selected at random, then the probability is 12/95.
1 in 52
0.3456
1/365 = 0.00274
random sample
1 out of 3600
The probability is 10 percent.
If the balls are selected at random, then the probability is 12/95.
1 in 52
The probability that it contains exactly 3 balls is 6/45 = 0.133... recurring.
The probability that exactly 4 out of 6 randomly selected vehicles will pass the test when the pass rate is 80% is approx 0.2458 (or nearly a quarter).
0.3456
1/365 = 0.00274
It depends on what the random variable is, what its domain is, what its probability distribution function is. The probability that a randomly selected random variable has a value between 40 and 60 is probably quite close to zero.
probability = 2/7 to be exact, 28/97 (about 28.87%)
The probability of selecting a 17 (or any number for that matter) is 1/20 or .05 or 5%.
random sample