Math, asked by epegg, 4 months ago

A bag contains 5 red marbles, 4 green marbles and 1 blue marble. A marble is chosen at random from the bag and not replaced; then a second marble is chosen. What is the probability both marbles are green?


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Answers

Answered by Anonymous
4

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Answered by Qwparis
0

The correct answer is \frac{2}{15}.

Given: Number of red marbles = 5.

Number of green marbles = 4.

Number of blue marbles = 1.

To Find: The probability that both marbles are green.

Solution:

Probability = \frac{Favorable outcome}{Total outcome}

Total marbles = green + blue + red

Total marbles = 4+1+5 = 10

Probability that marble chosen is green P(G) = \frac{Green marbles}{Total marbles}

P(G) = \frac{4}{10}=\frac{2}{5}.

Now, if green ball is chosen without replacement.

Total balls = 10-1=9

Green balls = 4-1=3

P(G) = \frac{Green marbles}{Total marbles}

P(G) = \frac{3}{9} = \frac{1}{3}

So, the probability both marbles are green = (Initial probability)(probability after choosing first marble without replacement)

=  \frac{2}{5}*\frac{1}{3}=\frac{2}{15}.

Hence, the probability both marbles are green is \frac{2}{15}.

#SPJ2

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