Problem 1
Answer: 5/14
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Step-by-step explanation:
We have 5 males and 3 females, so there are 5+3 = 8 mice total. The probability of selecting a male is 5/8. After that male is chosen, there are 5-1 = 4 of them left out of 8-1 = 7 total. This subtraction happens because the first mouse selected is not replaced. The probability of picking another male is 4/7.
Multiplying these fractions leads to the answer
(5/8)*(4/7) = (5*4)/(8*7) = 20/56 = 5/14
Note: to reduce 20/56 into 5/14, you divide both parts by the GCF 4.
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Problem 2
Answer: 9/14
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Step-by-step explanation:
The events "selecting two males" and "selecting at least one female" are complementary events. One or the other, but not both, must happen.
If we let
- A = event of selecting two males
- B = event of selecting at least one female (ie one or more females)
then we can say
P(A) + P(B) = 1
P(B) = 1 - P(A)
P(B) = 1 - (5/14)
P(B) = (14/14) - (5/14)
P(B) = (14-5)/14
P(B) = 9/14
Note how P(A) = 5/14 is the result from problem 1.
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Problem 3
Answer: 15/28
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Step-by-step explanation:
The probability of selecting a male is 5/8. After the male is chosen, there are 8-1 = 7 mice left. The probability the second mouse is female is 3/7 since there are 3 female mice out of 7 left total.
We get this result after multiplying the values: (5/8)*(3/7) = 15/56
Now consider that the first mouse is female. The probability of this is 3/8. The probability that the second mouse is male is 5/7.
The fractions multiply to (3/8)*(5/7) = 15/56. We get the same result as before due to the numerators swapping places, but not much else happens. So we have a sort of symmetry going on here.
The last step is to add the two results we got:
(15/56)+(15/56) = (15+15)/56 = 30/56 = 15/28
Or you could compute it like this
2*(15/56) = (2*15)/56 = 30/56 = 15/28
The '2' is to signify there are two ways to select exactly one male and exactly one female, where the order doesn't matter.
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Here's another way we can get the answer.
The probability we get both males is 5/14 found back in problem 1.
The probability we get both females will follow the same idea as problem 1, and we would compute (3/8)*(2/7) = 6/56 = 3/28
Add those fractions: (5/14) + (3/28) = (10/28) + (3/28) = 13/28
Now consider the events
- C = event of selecting two mice of the same gender (both are male OR both are female; pick 1 scenario only)
- D = event of selecting two mice of different genders (the first is male, the other is female, or vice versa)
Events C and D are complementary events.
We calculated that P(C) = 13/28, which means
P(D) = 1 - P(C)
P(D) = 1 - (13/28)
P(D) = (28/28) - (13/28)
P(D) = (28-13)/28
P(D) = 15/28