Question

A box of art supplies include the following: 5 markers - a blue, orange, green, and 2 red6 crayons - a red, orange, yellow, green, blue, and purple 4 colored pencils - red, orange, yellow, and greenIf two items are randomly taken from the box without replacement, which combinations will have a probability of 1/7? select all that apply.A.) choosing a crayon first and then a colored pencilB.) choosing a marker first and then a colored pencil C.) choosing a crayon first and then a marker.D.) choosing a crayon and then choosing another crayon

Answer 1

Let's calculate the probability for each case.

A) crayon first then colored pencil

We have 6 crayons among 15 items, so the probability is 6/15.

Then, we will have 4 colored pencils among 14 items (one crayon removed), so the probability is 4/14.

The final probability is the product of these two probabilities:

$P=\frac{6}{15}\cdot\frac{4}{14}=\frac{2}{5}\cdot\frac{2}{7}=\frac{4}{35}$

B) Marker and then colored pencil.

The probability for the marker first is 5/15, and the probability for the colored pencil after is 4/14, so:

$P=\frac{5}{15}\cdot\frac{4}{14}=\frac{1}{3}\cdot\frac{2}{7}=\frac{2}{21}$

C) crayon and then marker.

The probability for the crayon first is 6/15, and the probability for the marker after is 5/14, so:

$P=\frac{6}{15}\cdot\frac{5}{14}=\frac{2}{5}\cdot\frac{5}{14}=2\cdot\frac{1}{14}=\frac{1}{7}$

D) crayon and then another crayon

The probability for the crayon first is 6/15, and the probability for the second crayon is 5/14 (we have one less crayon for the second pick, so only 5 crayons among 15 items), so:

$P=\frac{6}{15}\cdot\frac{5}{14}=\frac{2}{5}\cdot\frac{5}{14}=2\cdot\frac{1}{14}=\frac{1}{7}$

So the correct options are C and D.