There are 6 apples, 8 oranges, and 10 peaches in an ice chest. If Maury gets to pick twice and he puts back what he selected the first time, what is the probability that he will get two peaches?

Hello!

Since Maury replaces what she takes out the probability will always be x/24, as 24 is the total number of fruits. We need to find the probability of getting one peach first.

10/24=5/12

Since she replaces the peach, she has a 5/12 chance again. We will multiply our probabilities.

5/12(5/12)=25/144, which is about 17.36%.

I hope this helps!

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kanest kanest · Answer 2 · 2018-06-29T23:22:05+02:00

The probability of Maury picking a peach is given by the following fraction:

[tex]\frac{\text{number of peaches}}{\text{total amount of fruit}}[/tex]

There are 10 peaches. To find the total amount of fruit, add the apples, oranges, and peaches together:

[tex]6 + 8 + 10 = 24[/tex]

The probability of Maury picking a peach will be the following fraction:

[tex] \frac{10}{24} [/tex]

This can be simplified by dividing the numerator and denominator by their GCF (greatest common factor):

Factors of 10: {1,2,5,10}
Factors of 24: {1,2,3,4,6,8,12,24}

GCF = 2

[tex]\frac{10}{24} \div \frac{2}{2} = \frac{5}{12}[/tex]

There is a 5/12 chance of Maury picking a peach.

Maury puts back whatever he chooses into the chest. Because the probability asks for the chance of Maury picking two peaches, we can set up the following equation:

[tex] \frac{5}{12} \times \frac{5}{12} = \frac{25}{144}[/tex]

There is a 25/144 chance of Maury picking two peaches, or a 17.36% chance.