Question

A bag contains 6 red marbles, 6 white marbles, and 4 blue marbles. What is the probability that someone could pull a white marble from the bag, keep it, and then pull a red marble on the second try?

Answer 1

Answer:

The probability of getting white marble in the first try and red marble in the second try (without replacement) is $\frac{3}{20}$

Step-by-step explanation:

Given: A bag contains 6 red marbles, 6 white marbles, and 4 blue marbles.

The total number of marbles in the bag = 6 + 6 + 4 = 16

Someone is pull a marble.

The probability of getting a white marble P(W)= $\frac{The number of favorable outcomes}{The total number of possible outcomes}$

P(W) = $\frac{6}{16}$

Then that marble kept it (with out replacement).

Now the total number of marbles in the bag = 6(red) + 5(white) + 4(blue) = 15

In the second try, the probability of getting a red marble P(R) = $\frac{6}{15}$

These two events are dependent, because the out comes of the first event influences the outcome of the second event.

So, the probability that someone could pull a white marble from the bag, keep it, and then pull a red marble on the second try = P(W) *P(R)

= $\frac{6}{16} *\frac{6}{15}$

Now we can multiply the fractions and simplify.

= $\frac{36}{240}$

Here GCF of 36 and 240 is 12. So divide both the numerator and the denominator by 12, we get

= $\frac{3}{20}$

So, the probability of getting white marble in the first try and red marble in the second try (without replacement) is $\frac{3}{20}$