An experiment consists of drawing 10 cards from an ordinary 52-card deck. recall this deck consists cards of 13 different denominations: a, 2-10, j, q, k and four suits if the drawing is made with replacement, \ if the drawing is made without replacement, find the probability that at least 9 cards will have the same suit.

This will be the sum of the probability that exactly 10 have the same suit plus the probability that 9 of 10 have the same suit.

With replacement first. Let's focus on a single suit, say hearts.

The probability of drawing on a heart on the first draw is 1/4. Since we're replacing, that's the probability of a heart for cards two through ten. So the probability of all hearts is .

Since there are four suits, the probability of drawing all the same suit is four times the above, .

The probability of drawing the first 9 hearts then a different suit for the 10th is .

The different suit may be any of the cards so we multiply this by 10 to get the probability of 9 of 10 hearts: .

We can have any suit, so we multiply this by a factor of 4, giving .

The probability we seek is the sum P(10 the same) + P(9 the same)

P(with replacement) = 0.012%

Now let's do it again without replacement.

P(Exactly 10 hearts). We have a 13/52 chance of picking a heart on the first card. Given we did, we have a 12/51 chance of picking a heart on the second card. So for all 10, we get

Let's write that as

Again the four suits gives us

For 9 of 10, the last fraction 4/43 becomes 39/43, then we multiply by 10 because any of the cards might not match and 4 to cover the suits.

So a total probability of

P(without replacement) = 0.00071%

A lot smaller.

I'd have to check this pretty hard to be sure it was right, which I haven't done.

4x and 6y

im not sure though

answer:

what is your question that you ask

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