Mathematics, 10.01.2020 06:31 fonsworth5

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

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Answer from: lisacarter0804

a. $\dfrac{1}{7}$

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) $=\dfrac{\text{No. of blue marbles}}{\text{Total marbles}}$

$=\dfrac{6}{21}=\dfrac{2}{7}$

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) = $\dfrac{\text{No. of red marbles}}{\text{remaining marbles }}$

$=\dfrac{10}{20}=\dfrac{1}{2}$

∵ the events of pulling marbles independent .

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

$=\dfrac{2}{7}\times\dfrac{1}{2}=\dfrac{1}{7}$

Hence, The probability of pulling out one blue marble and then one red marble without replacing the first blue marble = $\dfrac{1}{7}$

∴ Correct answer is a. $\dfrac{1}{7}$ .

Answer from: Tirone

a. $\frac{1}{7}$

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.

$P(\text{Blue})=\frac{\text{Number of blue marbles}}{\text{Total marbles}}$

$P(\text{Blue})=\frac{6}{10+5+6}$

$P(\text{Blue})=\frac{6}{21}$

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.

$P(\text{Red})=\frac{\text{Number of Red marbles}}{\text{Total marbles}}$

$P(\text{Blue})=\frac{10}{20}$

$P(\text{Blue})=\frac{1}{2}$

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

$P(\text{Red and Blue})=\frac{6}{21}\times \frac{1}{2}$

$P(\text{Red and Blue})=\frac{6}{42}$

$P(\text{Red and Blue})=\frac{1}{7}$

Therefore, the probability of pulling out one blue marble and then one red marble without replacing the first blue marble is $\frac{1}{7}$ and option 'a' is the correct choice.

Answer from: Quest

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Answer from: Quest

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