Question

Bag A contains 9 red marbles and 3 green marbles. Bag B contains 9 black marbles and 6 orange marbles, Find the probability of selecting one green marble from bag A and one black marble from bag B.

Answer 1

Answer:

3/20

Step-by-step explanation:

Prob. (For green (Bag A)) = 3/12

Prob. (For Black (Bag B)) = 9/15

3/12 x 9/15 = 3/20

Answer 2

The probability of selecting one green marble from bag A and one black marble from bag B is 0.15

There are 3 green marbles in bag A out of a total of 12 marbles (i.e. 3 green and 9 red).

So, the probability of selecting a green marble from bag A is:

$Pr = \frac 3{12}$

There are 9 black marbles in bag B out of a total of 15 marbles (i.e. 9 black and 6 orange).

So, the probability of selecting a black marble from bag B is:

$Pr = \frac 9{15}$

The required probability is then calculated as:

$P = \frac 3{12} * \frac 9{15}$

Multiply

$P = 0.15$

Hence, the probability of selecting one green marble from bag A and one black marble from bag B is 0.15

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