Question

Aptitude
Question 15 of 20
There are 12 black, 16 blue and 8 red pens in a bag. Three pens are drawn from the bag without replacement. What is the probability that the
first pen is blue, the second pen is red and the third pen is black?
O 86/1875
O 36/1290
53/1260
O 64/1785​

Answer 1

Answer:

i think the right is 1/36

Step-by-step explanation:

Answer 2

Answer:

The probability  that the  first pen is blue, the second pen is red and the third pen is black is $\frac{64}{1785}$

Step-by-step explanation:

Given the number of black (Bk) pens = 12

Number of blue (B) pens = 16

Number of red (R) pens = 8

Total number of pens = 12 + 16 + 8 = 36

3 pens are drawn without replacement.

We need to find the probability of B R Bk in their respective picks.

We know that probability = $\frac{Favourable outcomes}{Total number of outcomes}$

First of all, the probability of the first pen taken out to be blue = 16/36

Then, the probability of the second pen taken out to be red = 8/35

(Total number of outcomes become 35 because, 1 pen is already taken out Hence, 36 - 1 = 35)

Then, the probability of the third pen taken out to be black = 12/34

(Total number of outcomes become 34 because, 2 pens are already taken out Hence, 36 - 2 = 34)

Therefore, the probability  that the  first pen is blue, the second pen is red and the third pen is black is $\frac{16}{36}$ * $\frac{8}{35}$ * $\frac{12}{34}$ = $\frac{64}{1785}$