SOLUTION: You are dealt two cards successively without replacement from a standard deck of 52 playing cards. Find the probability that the first card is a two and the second card is a ten. R

Question 73016This question is from textbook Elementary Statistics
: You are dealt two cards successively without replacement from a standard deck of 52 playing cards. Find the probability that the first card is a two and the second card is a ten. Round your answer to three decimal places. This question is from textbook Elementary Statistics

Found 2 solutions by jim_thompson5910, bucky:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Probability that the first card is a two: There are 4 two cards out of 52
P(card of two)=4/52=1/13
Probability that the second card is a ten: There are 4 ten cards out of 51 (take one 2 out of the deck)
P(card of ten)=4/51
Now since these two events need to happen at once, we multiply the two probabilities
P(two and a ten)=P(card of two)*P(card of ten)=(1/13)*(4/51)=4/663
So the chances of drawing a 2 and then a ten is 4/663 or 0.00603 (a 0.6% chance)
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
There are 4 twos in a deck of 52 cards. So the chance that the first card is a two is 4 in 52
which is the same as 4 divided by 52 and this is a probability of 0.076923076.
.
Once the first card is dealt, there are 51 cards remaining in the deck. In that group there
are 4 tens. So the chances of drawing a 10 are 4 in 51 which is the same as 4 divided by
51 and this is 0.078431372.
.
The chance of drawing first a two and then a ten is the product of these two probabilities
and this is:
.
(0.076923076)*(0.078431372) = 0.006033182389
.
that's your answer. You can round it off. It translates that if you tried this dealing
(two cards from a deck of 52 without putting the card back that you drew first) a thousand times
the odds are that you would first draw a two and then a ten about 6 times.
.
Hope this helps you to see how to do this problem.

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