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Printjmc
counting and probability senior
Problem
Three cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is a 4, the second card is a , and the third card is a 2?
Solution
There are 4 exclusive cases:
Case 1: first card not a and second card not a 2.
There are 3 cards that are 4's but not a , so the probability for the first card is . Next, there are 12 s remaining that aren't a 2, so the probability for the second card is . Finally, there are four 2's remaining, so the probability for the third card is . Hence, this case gives a probability of . (We leave the fraction in these terms rather than reducing because we know that we're going to need to add fractions later.)
Case 2: first card not a and second card the 2.
There are 3 cards that are 4's but not a , so the probability for the first card is . Next, there is only one 2, so the probability for the second card is . Finally, there are three 2's remaining, so the probability for the third card is . Hence, this case gives a probability of .
Case 3: first card the 4 and second card not a 2.
There is only one 4, so the probability for the first card is . Next, there are 11 s remaining that aren't a 2, so the probability for the second card is . Finally, there are four 2's remaining, so the probability for the third card is . Hence, this case gives a probability of .
Case 4: first card the 4 and second card the 2.
There is only one 4, so the probability for the first card is . Next, there is only one 2, so the probability for the second card is . Finally, there are three 2's remaining, so the probability for the third card is . Hence, this case gives a probability of .
So the overall probability is .
Case 1: first card not a and second card not a 2.
There are 3 cards that are 4's but not a , so the probability for the first card is . Next, there are 12 s remaining that aren't a 2, so the probability for the second card is . Finally, there are four 2's remaining, so the probability for the third card is . Hence, this case gives a probability of . (We leave the fraction in these terms rather than reducing because we know that we're going to need to add fractions later.)
Case 2: first card not a and second card the 2.
There are 3 cards that are 4's but not a , so the probability for the first card is . Next, there is only one 2, so the probability for the second card is . Finally, there are three 2's remaining, so the probability for the third card is . Hence, this case gives a probability of .
Case 3: first card the 4 and second card not a 2.
There is only one 4, so the probability for the first card is . Next, there are 11 s remaining that aren't a 2, so the probability for the second card is . Finally, there are four 2's remaining, so the probability for the third card is . Hence, this case gives a probability of .
Case 4: first card the 4 and second card the 2.
There is only one 4, so the probability for the first card is . Next, there is only one 2, so the probability for the second card is . Finally, there are three 2's remaining, so the probability for the third card is . Hence, this case gives a probability of .
So the overall probability is .
Final answer
\frac{1}{663}