Question

A bag contains 4 red marbles, 8 blue marbles and 2 green marbles. If two marbles are drawn out of the bag, what is the probability, to the nearest 10th of a percent, that both marbles drawn will be red?

Answer 1

Answer:

The probability, to the nearest 10th of a percent, that both marbles drawn will be red is 28.5%

Step-by-step explanation:

Number of red marbles = 4

Number of blue marbles = 8

Number of Green marbles = 2

Total number of marbles = 4+8+2 =14

We need to find the probability, to the nearest 10th of a percent, that both marbles drawn will be red?

The formula used will be: $Probability=\frac{Number\:of\:favourable\:outcomes}{Number\:of\;possible\:outcomes}$

In our case, Number of favourable outcomes = 4 (as red marbles are drawn)

Number of possible outcome = 14 (Total number of marbles)

The probability will be:

$Probability=\frac{Number\:of\:favourable\:outcomes}{Number\:of\;possible\:outcomes}\\\\Probability=\frac{4}{14}\\ Probability=0.2857$

Now, calculating nearest 10th of a percent

$Probability = 0.2857*100\\Probability=28.5%$

So, the probability, to the nearest 10th of a percent, that both marbles drawn will be red is 28.5%