Question

Malik has a bag with 7 green marbles and 3 black marbles.

Answer 1

1) Probability: 7/30 (or 23.3%)

2) Probability: 21/100 (or 21.0%)

Step-by-step explanation:

1)

At the beginning, there are 10 marbles in total in the bag, of which 7 are green and 3 are blacks.

So the probability of choosing a green marble as first is:

$p(g)=\frac{7}{7+3}=\frac{7}{10}$

Then, after he picked a green marble, there are only 9 marbles left in the bag, of which 6 are green and 3 are black. Therefore, at the second time, the probability of choosing a black marble will be

$p(b)=\frac{3}{3+6}=\frac{3}{9}=\frac{1}{3}$

Therefore, the probability that Malik will draw a green marble first and then a black marble is the product of the two probabilities:

$p(gb)=p(g)p(b)=\frac{7}{10}\cdot \frac{1}{3}=\frac{7}{30}$

In percentage, this is $\frac{7}{30}\cdot 100 =23.3\%$

2)

In this second case, the same experiment is repeated by this time the first ball is put back into the back after the  first draw.

The probability that Malik will draw a green marble at the first attempt is the same as before:

$p(g)=\frac{7}{7+3}=\frac{7}{10}$

Later, the green marble is put back in the bag, so we still have 7 green marbles and 3 black marbles. Therefore, the probability of choosing a black marble at the second draw is

$p(b)=\frac{3}{3+7}=\frac{3}{10}$

Therefore, the overall probability is

$p(gb)=p(g)p(b)=\frac{7}{10}\cdot \frac{3}{10}=\frac{21}{100}$

In percentage, $\frac{21}{100}\cdot 100 = 21.0\%$

3)

The probability in the second case is lower because in the second case, the green marble is put back into the bag after the first draw, therefore at the second draw the probability of choosing a black marble is less than the first case (because in the 2nd case, there are more marbles in total to choose from, so the probability of choosing a black marble will be less).

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