A bag of peanuts in their shells contains 181 peanuts. 65 of the shells contain one peanut, 111 of the shells contain two peanuts, and the rest contain three peanuts. Of the shells with one peanut, 16 of them are cracked. Of the shells with two peanuts, 30 of them are cracked. None of the shells with three peanuts are cracked. One peanut shell is randomly selected from the bag.
A. What is the probability the shell is cracked?B. What is the probability the shell is not cracked and it contains two peanuts?C. What is the probability the shell is not cracked or it contains two peanuts?D. What is the probability the shell is not cracked given that it contains two peanuts?

To calculate the probability of the shell bein cracked you have to add all possible cases in which it is cracked and divide it by the total of peanuts in the bag:

[tex]P(Cracked)= \frac{(16+30)}{181}= 0.25[/tex]

B. What is the probability the shell is not cracked (Cr) and it contains two peanuts(2p)?

This probability is an intersection P(Cr∩2p), to calculate it you have to divide the total of peanuts that fulfill these characteristics and divide it by the total number of peanuts in the bag.

[tex]P(Crn2p)= \frac{30}{181}= 0.17[/tex]

C. What is the probability the shell is not cracked (Cr') or it contains two peanuts(2p)?

This probability is the union between two events, the event "cracked" and the event "two peanuts", symbolically: P(Cr'∪2p), these two events are not mutually exclusive, this means that they can happen at the same time, you have to apply the following formula to calculate it:

[tex]P(Cr'u2p)= P(Cr') + P(2p) - P(Cr'n2p)[/tex]

Since these events aren't mutually exclusive, some of them fulfill both categories, this means they are counted when you calculate the probability of "not cracked" and again when you calculate the probability of "2 peanuts" therefore you need to subtract their intersection to obtain the correct probability.

[tex]P(Cr'u2p)= \frac{(49+81+5)}{181} + (\frac{111}{181} ) - (\frac{81}{181} )= 0.76+0.61-0.45=0.92[/tex]

D. What is the probability the shell is not cracked given that it contains two peanuts?

This is a conditional probability, this means that both events are dependant and the occurrence of "2p" modifies the probability of the peanut shell to be "Cr'", symbolically: P(Cr'/2p) and you calculate it using the following formula:

[tex]P(Cr'/2p)= \frac{P(Cr'n2p)}{P(2p)}= \frac{0.45}{0.61}= 0.74[/tex]

I hope it helps!

francocanacari francocanacari · Answer 2 · 2021-11-23T11:57:31+01:00

A) The probability the shell is cracked is 25.4%.

B) The probability the shell is not cracked and it contains two peanuts is 44.75%.

C) The probability the shell is not cracked or it contains two peanuts is 91.16%.

D) The probability the shell is not cracked given that it contains two peanuts is 72.97%.

Since a bag of peanuts in their shells contains 181 peanuts. 65 of the shells contain one peanut, 111 of the shells contain two peanuts, and the rest contain three peanuts, and, of the shells with one peanut, 16 of them are cracked, and of the shells with two peanuts, 30 of them are cracked, while none of the shells with three peanuts are cracked, and one peanut shell is randomly selected from the bag, to determine A) what is the probability the shell is cracked; B) what is the probability the shell is not cracked and it contains two peanuts; C) what is the probability the shell is not cracked or it contains two peanuts; and D) what is the probability the shell is not cracked given that it contains two peanuts; the following calculations must be performed:

A)

181 = 100
16 + 30 = X
46 x 100/181 = X
25.4 = X

B)

181 = 100
111 - 30 = X
81 x 100 / 181 = X
44.75 = X

C)

181 = 100
111 + 5 + 49 = X
165 x 100 / 181 = X
91.16 = X

D)

111 = 100
111 - 30 = X
81 x 100 / 111 = X
72.97 = X

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