Abag contains 10 red, 5 yellow, and 6 blue marbles. what is the probability of pulling out one blue marble and then one red marble without replacing the first blue marble?

a.17
b.415
c.160
d.1621

 a.

Step-by-step explanation:

Given : A bag contains 10 red, 5 yellow, and 6 blue marbles.

Total marble = 10+5+6= 21

Probability of pulling one blue marble first :  P(First blue)

Also, the second marble is drawn without replacement , so after drawing one marble , the remaining marbles = 21-1=20

Then P(Second marble is red marble) =

∵ the events of pulling marbles independent .

⇒ P( Blue then red)= P(First blue) x P(Second marble is red)

Hence, The probability of pulling out one blue marble and then one red marble without replacing the first blue marble =

∴ Correct answer is a. .

a.

Step-by-step explanation:

We have been given that a bag contains 10 red, 5 yellow, and 6 blue marbles. We are asked to find the probability of pulling out one blue marble and then one red marble without replacing the first blue marble.

Let us find probability of pulling out a blue marble.

Since one blue marble is pulled out and it is not replaced, so total marbles will be 20.

Now, let us find probability of pulling out a red marble.

To find the probability of pulling out one blue marble and then one red marble without replacement, we will multiply both probabilities as:

Therefore, the probability of pulling out one blue marble and then one red marble without replacing the first blue marble is and option 'a' is the correct choice.

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