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A cookie jar contains 3 chocolate chip cookies, 4 oatmeal raisin cookies, and 5 sugar cookies.

Event [tex]\( A \)[/tex] = drawing a chocolate chip cookie on the first draw
Event [tex]\( B \)[/tex] = drawing an oatmeal raisin cookie on the second draw

If two cookies are drawn from the jar, one after the other and not replaced, what is [tex]\( P(A \text{ and } B) \)[/tex] expressed in simplest form?

A) [tex]\(\frac{1}{15}\)[/tex]

B) [tex]\(\frac{1}{11}\)[/tex]

C) [tex]\(\frac{1}{4}\)[/tex]

D) [tex]\(\frac{4}{11}\)[/tex]



Answer :

GinnyAnswer
Let's solve this problem step-by-step to determine the probability of drawing a chocolate chip cookie first (Event A) and then an oatmeal raisin cookie second (Event B).

1. Calculate the Probability of Event A:

Event [tex]$A$[/tex] is the event of drawing a chocolate chip cookie from the jar on the first draw. The cookie jar contains:

- 3 chocolate chip cookies
- 4 oatmeal raisin cookies
- 5 sugar cookies

Therefore, the total number of cookies in the jar is:
[tex]\[ 3 + 4 + 5 = 12 \][/tex]

The probability of drawing a chocolate chip cookie on the first draw is:
[tex]\[ P(A) = \frac{\text{Number of chocolate chip cookies}}{\text{Total number of cookies}} = \frac{3}{12} = \frac{1}{4} \][/tex]

2. Calculate the Probability of Event B given Event A:

Event [tex]$B|A$[/tex] is the event of drawing an oatmeal raisin cookie on the second draw, given that the first cookie drawn was a chocolate chip cookie.

After drawing one chocolate chip cookie, there are:

- 2 chocolate chip cookies left
- 4 oatmeal raisin cookies
- 5 sugar cookies

This means there are now a total of:
[tex]\[ 12 - 1 = 11 \text{ cookies remaining} \][/tex]

The probability of drawing an oatmeal raisin cookie after one chocolate chip cookie has been drawn is:
[tex]\[ P(B|A) = \frac{\text{Number of oatmeal raisin cookies}}{\text{Remaining total number of cookies}} = \frac{4}{11} \][/tex]

3. Calculate the Joint Probability [tex]$P(A \text{ and } B$[/tex]):

The probability of both events happening (drawing a chocolate chip cookie first and then an oatmeal raisin cookie) is calculated by multiplying the probability of Event [tex]$A$[/tex] by the conditional probability of Event [tex]$B$[/tex] given Event [tex]$A$[/tex]:
[tex]\[ P(A \text{ and } B) = P(A) \times P(B|A) \][/tex]
[tex]\[ P(A \text{ and } B) = \frac{1}{4} \times \frac{4}{11} = \frac{4}{44} = \frac{1}{11} \][/tex]

Therefore, the probability of drawing a chocolate chip cookie on the first draw followed by an oatmeal raisin cookie on the second draw is [tex]\(\frac{1}{11}\)[/tex].

The correct answer is:

B) [tex]\(\frac{1}{11}\)[/tex]

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