A pack of 36 cards includes 20 numbered cards from 6 to 10 inclusive, 4 aces and 12 picture cards. If a hand of 5 cards is selected at random, how do you find the probability of receiving at least 2 aces?
31,776
Let's first check to see if I can say this if I do this:
A hand with zero aces plus one ace plus two aces plus three aces plus four aces equals all possible hands.
Therefore, there are two ways we can work this out: either we find all possible hands and subtract out hands with 0 and 1 ace, or we can add up the hands with 2, 3, and 4 aces. Since both ways require the same amount of work, I'll go with the more interesting subtraction method.
There are 36 combinations of possible hands, so choose option number five.
The number of hands with zero aces indicates that we have five cards left out of the 32 total cards and zero aces out of the four available:
Furthermore, the number of hands with an ace indicates that we have four cards remaining in the deck, and one ace out of those four:
This results in:
Given that this may appear to be a relatively small number in comparison to our current findings, let's calculate the hands that contain 2, 3, and 4 aces:
2 aces:
3 aces:
4 aces:
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To find the probability of receiving at least 2 aces in a hand of 5 cards selected at random from the pack, you can use the concept of combinations.
 First, determine the total number of ways to select a hand of 5 cards from the pack of 36 cards. This can be calculated using combinations, denoted as ( C(n, k) ), which gives the number of combinations of selecting ( k ) items from a set of ( n ) items.
[ C(36, 5) = \frac{36!}{5!(365)!} ]
 Next, determine the number of ways to select exactly 2 aces from the 4 aces available in the pack. This can be calculated as:
[ C(4, 2) = \frac{4!}{2!(42)!} ]
 Determine the number of ways to select the remaining 3 cards from the remaining cards in the pack, excluding the aces.
[ C(32, 3) = \frac{32!}{3!(323)!} ]

Calculate the number of favorable outcomes where you have at least 2 aces by multiplying the number of ways to choose exactly 2 aces and the number of ways to choose the remaining 3 cards.

Finally, divide the number of favorable outcomes by the total number of possible outcomes to find the probability.
[ P(\text{at least 2 aces}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} ]
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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