Question

Plz read and answerrrrrrrrrrr

Answer 1

Answer:

16.7%.

Step-by-step explanation:

There are initially $24 + 15 + 8 = 47$ pencils in the bag.

Take a pencil out of this bag of 47 pencils. 15 out of the 47 pencils blue. Let $A$ represent the event of getting a blue pencil on the first pick. The probability of getting a blue pencil is:

$\displaystyle P(A) = \frac{15}{47}$.

There are now $47 - 1 = 46$ pencils left in the bag. However, given that the first pencil removed from the bag is blue, the number of red pencils in the bag will still be 24. Take another pencil out of this bag of 46 pencils. Let $B$ represent the event of getting a red pencil on the second pick. The possibility that the second pencil is red given that the first pencil is blue will be:

$\displaystyle P(B|A) = \frac{24}{46}$.

The question is asking for the possibility that the first pencil is blue and the second pencil is red. That is:

$\displaystyle P(A\cap B) = P(A) \cdot P(B|A) = \frac{15}{47}\times \frac{24}{46} = 0.166512 \approx 16.7\%$.