Question

check my answer please?

Answer 1

a basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5 peaches. Jameson picks a fruit at random and does not replace it. Then Brittany wingspans a fruit at random. What is the probability that Jameson gets a banana and Brittany gets a pear?

A. 4/27
B. 1/49
C. 2/91
D. 27/91

I think A

Answer 2

please help

Answer 3

Notice the key phrase in this question, the fact that Jameson picks a fruit but DOES NOT REPLACE IT, that means, after he takes out the fruit, that fruit is no longer in the basket for the next trial. So similar to your last question, to calculate the probability of an intersection of events\[P(AandB)=P(A)*P(B|A)\]where in this case, P(A) is the probability that Jameson gets a banana, while P(B) is the probability that Brittany gets a pear after Jameson already took one banana out.

Side Note:\[P(B|A)\]means the probability that the event B occurs given, or after we know that event A has occurred.

Hope that helps out

Answer 4

so it would be P = 2/14 * 2/13 ? how do i solve that?

Answer 5

would it be C?

Answer 6

Exactly! Your result will be\[\frac{ 2 }{ 14 }*\frac{ 2 }{ 13 }=\frac{ 2 }{ 7*13 }\]since\[\frac{ 2 }{ 14 }=\frac{ 1 }{ 7 }\]and you are doing great since you know the answer is C
:D keep up the good work.