what is the probability of a six card hand cantoning no fives?

probability of one 5 = 4/52

probability of two 5s = 4/52*3/51
probability of three 5s =4/52*3/51*2/50
probability of four 5s=4/52*3/51*2/50*1/49
so probability of at least one 5 = sum of all those
subtract that from 1 to get the probability of none.

or

number of six card hands = C(52,6)
number of six card hands with no fives = C(48,6)

prob of your event = C(48,6)/C(52,6) = .60277

To calculate the probability of a six-card hand containing no fives, we need to know the total number of possible six-card hands and the number of favorable outcomes (hands with no fives).

Step 1: Determine the total number of possible six-card hands:
In a standard deck of 52 playing cards, there are 52 choices for the first card, 51 choices for the second card, 50 choices for the third card, and so on. Since we are dealing with a six-card hand, we multiply these numbers together to obtain the total number of possible outcomes: 52 * 51 * 50 * 49 * 48 * 47 = 311,875,200.

Step 2: Determine the number of favorable outcomes:
To have no fives in the hand, we need to choose from the remaining 48 cards that are not fives. Since there are four fives in the deck, out of the 52 cards, there are 52 - 4 = 48 choices for the first card, 47 choices for the second card, 46 choices for the third card, and so on. We multiply these numbers together to obtain the number of favorable outcomes: 48 * 47 * 46 * 45 * 44 * 43 = 28,475,013,024.

Step 3: Calculate the probability:
The probability of an event is given by dividing the number of favorable outcomes by the total number of possible outcomes. Therefore, the probability of a six-card hand containing no fives is:

P(no fives) = favorable outcomes / total outcomes
P(no fives) = 28,475,013,024 / 311,875,200
P(no fives) ≈ 0.0913

So, the probability of a six-card hand containing no fives is approximately 0.0913, or 9.13%.