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?

Answer 1

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.

#((36),(5))=(36!)/((31!)(5!))=(36xx35xx34xx33xx32)/120="376,992"#

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:

#((4),(0))((32),(5))=(1)(32xx31xx30xx29xx28)/120="201,376"#

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:

#((4),(1))((32),(4))=(4)(32xx31xx30xx29)/24="143,840"#

This results in:

#"376,992"-"201,376"-"143,840"="31,776"#

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:

#((4),(2))((32),(3))=(6)(32xx31xx30)/6="29,760"#

3 aces:

#((4),(3))((32),(2))=(4)(32xx31)/2=1984#

4 aces:

#((4),(4))((32),(1))=(1)(32)=32#
#"29,760"+1984+32="31,776"#
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Answer 2

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.

  1. 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) C(n, k) , which gives the number of combinations of selecting k k items from a set of n n items.

C(36,5)=36!5!(365)!C(36, 5) = \frac{36!}{5!(36-5)!}

  1. 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)=4!2!(42)!C(4, 2) = \frac{4!}{2!(4-2)!}

  1. Determine the number of ways to select the remaining 3 cards from the remaining cards in the pack, excluding the aces.

C(32,3)=32!3!(323)!C(32, 3) = \frac{32!}{3!(32-3)!}

  1. 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.

  2. Finally, divide the number of favorable outcomes by the total number of possible outcomes to find the probability.

P(at least 2 aces)=Number of favorable outcomesTotal number of possible outcomesP(\text{at least 2 aces}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}

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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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