A bag contains $3$ red marbles, $5$ blue marbles and $8$ green marbles. If two marbles are drawn out of the bag, what is the probability, to the nearest $1$ oth of a percent, that both marbles drawn will be blue? $\newline$ Answer:

Full solution

Q. A bag contains

3

red marbles,

5

blue marbles and

8

green marbles. If two marbles are drawn out of the bag, what is the probability, to the nearest

1

oth of a percent, that both marbles drawn will be blue?

\newline

Answer:

Calculate Total Marbles: Determine the total number of marbles in the bag. $\newline$ Add the number of red, blue, and green marbles together. $\newline$ $3$ red + $5$ blue + $8$ green = $16$ total marbles.
Calculate Probability of Drawing: Calculate the probability of drawing one blue marble. Since there are $5$ blue marbles out of $16$ total marbles, the probability of drawing one blue marble is $\frac{5}{16}$ .
Calculate Probability of Second Draw: Calculate the probability of drawing a second blue marble after one has already been drawn. $\newline$ After drawing one blue marble, there are now $4$ blue marbles left and $15$ marbles in total. The probability of drawing a second blue marble is now $\frac{4}{15}$ .
Calculate Probability of Both Events: Calculate the probability of both events happening together (drawing two blue marbles in a row). $\newline$ Multiply the probabilities of each individual event to find the combined probability. $\newline$ $(\frac{5}{16}) \times (\frac{4}{15}) = \frac{20}{240} = \frac{1}{12}$ .
Convert Probability to Percentage: Convert the probability to a percentage. $\newline$ To convert a fraction to a percentage, multiply it by $100$ . $\newline$ $(1/12) \times 100 = 8.333\ldots$
Round Percentage: Round the percentage to the nearest tenth of a percent. $8.333\ldots\%$ rounds to $8.3\%$ when rounded to the nearest tenth of a percent.