For this problem, assume that youre dealt 5 cards from a standard .pdf
For all probabilities in this sample space, the denominator = Find t.pdf
0 gefällt mir
Aufrufe
For all probabilities in this sample space, the denominator = Find the following probabilities: Solution Denominator = 52C4 a) There are 4 aces and 4 kings. So P(2 aces and 2 kings) = ways to select 2 aces x ways to select 2 kings / total ways = 4C2 x 4C2 / 52C4 = 1.329 x 10^-4 b) There are 26 red cards. So, P(all cards are red) = ways of selecting 4 red cards out of 26 red cards/total ways = 26C4/52C4 = 0.0552 c) P(1 ace, 1 king , 1 queen and any other card) = ways of selecting 1 ace out of 4 aces x ways of selecting 1 king out of 4 kings x ways of selecting 1 queen out of 4 queens x ways of selecting remaining card/total ways = 4 x 4 x 4 x 49/52C4 = 0.0115 d) P(3 red and 1 black) = ways of selecting 3 red cards out of 26 red cards x ways of selecting 1 black cards out of 26 black cards/total ways = 26C3 x 26C1/52C4 = 0.2497.
For all probabilities in this sample space, the denominator = Find t.pdf
- For all probabilities in this sample space, the denominator = Find the following probabilities: Solution Denominator = 52C4 a) There are 4 aces and 4 kings. So P(2 aces and 2 kings) = ways to select 2 aces x ways to select 2 kings / total ways = 4C2 x 4C2 / 52C4 = 1.329 x 10^-4 b) There are 26 red cards. So, P(all cards are red) = ways of selecting 4 red cards out of 26 red cards/total ways = 26C4/52C4 = 0.0552 c) P(1 ace, 1 king , 1 queen and any other card) = ways of selecting 1 ace out of 4 aces x ways of selecting 1 king out of 4 kings x ways of selecting 1 queen out of 4 queens x ways of selecting remaining card/total ways = 4 x 4 x 4 x 49/52C4 = 0.0115 d) P(3 red and 1 black) = ways of selecting 3 red cards out of 26 red cards x ways of selecting 1 black cards out of 26 black cards/total ways = 26C3 x 26C1/52C4 = 0.2497