A box contains 11 marbles, 7 red, and 4 green. Five of these marbles are removed at random. If the probability of drawing a green marble is now 0.5, how many red marbles were removed?

fumefluosault7pa
2022-02-12
Answered

A box contains 11 marbles, 7 red, and 4 green. Five of these marbles are removed at random. If the probability of drawing a green marble is now 0.5, how many red marbles were removed?

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Finn Lucero

Answered 2022-02-13
Author has **17** answers

Given:

Red marbles$=7$

Green marbles$=4$

Total no of marbles$=11$

Number of Marbles removed$=5$

Marbles in the box$=11-5=6$

Probability of drawing one green marble is$=0.5$

Among the remaining 6 marbles, x number of marbles are green.

Then Probability of drawing one green marble is = number of green marbles / Total Number of marbles.

$P}_{1\text{}\text{green}}=\frac{x}{6$

We know${P}_{1\text{}\text{Green}}=0.5$

Then

$\frac{x}{6}=0.5$

$x=0.5\times 6=3$

After 5 marbles are removed from the box, the number of green marbles remaining is 3.

It means out of 4 green marbles, one green marble has been removed.

Out of 5 marbles one marble is green. Then the other four must be red.

Then number of red marbles removed were$=4$ .

Red marbles

Green marbles

Total no of marbles

Number of Marbles removed

Marbles in the box

Probability of drawing one green marble is

Among the remaining 6 marbles, x number of marbles are green.

Then Probability of drawing one green marble is = number of green marbles / Total Number of marbles.

We know

Then

After 5 marbles are removed from the box, the number of green marbles remaining is 3.

It means out of 4 green marbles, one green marble has been removed.

Out of 5 marbles one marble is green. Then the other four must be red.

Then number of red marbles removed were

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