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almond37 [142]
3 years ago
9

A box of Georgia peaches has 3 bad and 12 good peaches. (a) If you make a peach cobbler of 12 peaches randomly selected from the

box, what is the probability that there are no bad peaches in the peach cobbler? (b) What is the probability of having at least 1 bad peach in the peach cobbler? (c) What is the probability of having exactly 2 bad peaches in the peach cob- bler?
Mathematics
1 answer:
Eddi Din [679]3 years ago
3 0

Answer:

a) 0.21% probability that there are no bad peaches in the peach cobbler.

b) 99.79% probability of having at least 1 bad peach in the peach cobbler

c) 7.91% probability of having exactly 2 bad peaches in the peach cobbler.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the peaches are chosen is not important. So the combinations formula is used to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

(a) If you make a peach cobbler of 12 peaches randomly selected from the box, what is the probability that there are no bad peaches in the peach cobbler?

Desired outcomes:

12 good peaches, from a set of 12. So

D = C_{12,12} = \frac{12!}{12!(12 - 12)!} = 1

Total outcomes:

12 peaches, from a set of 15. So

T = C_{15,12} = \frac{15!}{12!(15 - 12)!} = 455

Probability:

p = \frac{D}{T} = \frac{1}{455} = 0.0021

0.21% probability that there are no bad peaches in the peach cobbler.

(b) What is the probability of having at least 1 bad peach in the peach cobbler?

Either there are no bad peaches, or these is at least 1. The sum of the probabilities of these events is 100%. So

p + 0.21 = 100

p = 99.79

99.79% probability of having at least 1 bad peach in the peach cobbler

(c) What is the probability of having exactly 2 bad peaches in the peach cob- bler?

Desired outcomes:

2 bad peaches, from a set of 3.

One good peach, from a set of 12.

D = C_{3,2}*C_{12,1} = \frac{3!}{2!(3-2)!}*\frac{12!}{1!(12 - 1)!} = 36

Total outcomes:

12 peaches, from a set of 15. So

T = C_{15,12} = \frac{15!}{12!(15 - 12)!} = 455

Probability:

p = \frac{D}{T} = \frac{36}{455} = 0.0791

7.91% probability of having exactly 2 bad peaches in the peach cobbler.

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Step-by-step explanation:

Given:

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We are given that m and n are both integers and we have to select all the statements that are true if m and n are also equal to each other.

(i) The given situation is: m - n = n - m

LHS = m - n

       = m - m   {because m and n are equal}

       = 0

RHS = n - m

       = n - n   {because m and n are equal}

       = 0

Hence, the given statement is true because LHS = RHS = 0.

(ii) The given situation is: +m (- n) = m - n

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RHS = m - n

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Hence, the given statement is true because LHS = RHS = 0.

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Hence, the given statement is not true because LHS \neq RHS.

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