# probability helpWatch

#1
could someone help me with these please..

1)what is the probability of drawing three tens in succession from a standard pack of 52 playing cards given that the cards are not replaced?

2)what is the probability of drawing four sevens in succession from a standard pack of 52 playing cards given that the cards are replced?

im thinking for example like this for question 1:

3/52 x 2/51 x 1/50 = X

is this the correct way to go about it?

thank u
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12 years ago
#2
Hi, for the first one:

There are four 10s to start with, so the probability you pick one is 4/52.
Then, there are three 10s left, so the probability is 3/51, etc.
so P(three tens in a row) = 4/52 * 3/51 * 2/50

Now you should be able to do question 2.
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#3

So can i just leave the answer as 4/52, 3/52, 2/50, or do i have to multiply it and get the answer for me to be actually answering the question?
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12 years ago
#4
(Original post by Kurdi_princess)

So can i just leave the answer as 4/52, 3/52, 2/50, or do i have to multiply it and get the answer for me to be actually answering the question?
You tell me! Have a read through your textbook, and try and work it out...
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12 years ago
#5
nothing like a maths question involving cards takes me back to poker
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12 years ago
#6
for the second 1:
zer are 4 sevens in a pack.given that the card is replaced,at any time before a draw,zer will be always 52 cards in the pack so that the prob. of obtaining 4 sevens consecutively will be:

(4/52)^4=1/28561
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12 years ago
#7
^ It would be better to work with 1/13 than 4/52. I would have much more trouble working out 52^4 and then trying to simplify the result than 13^4 and not having to simplify (even with a calculator it is still easier to use 1/13).

It's like when I saw someone in a non-calculator past GCSE paper trying to work out something like (12/17)/(6/34) (the numbers were actually quite a bit bigger, but this is just to illustrate) without simplifying so doing it like this:
(12/17)/(6/34)=(12/17)(34/6)=408/102=4
when they could have done:
(12/17)/(6/34)=(12/17)(34/6)=2*2=4
and avoided multiplying 12 by 34 (it's not that difficult, but in the real paper the numbers were bigger and the person was multiplying 3 digit numbers when he could have been using 1 digit numbers, so it took him about ten times as long as everyone else).

You may ask what the point of doing the last paragraph was. Basically I just want people who don't think to simplify the problem when dealing with fractions to see that it can be much easier than multiplying everything out.
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12 years ago
#8
(Original post by jahvins)
zer are 4 sevens in a pack.given that the card is replaced,at any time before a draw,zer will be always 52 cards in the pack so that the prob. of obtaining 4 sevens consequetively will be:
I love your accent... "consequetively" . I think I will take to writing in an accent...
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