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A box contains 8 red balls and 12 green balls. Two balls are chosen at random from the box and these balls are discarded (thrown away without observing their colour). Zuri later draws a ball from the box (of 18 balls). (i) Find the probability that Zuri draws a red ball (il) Given that Zuri draws a red ball, find the probability that the discarded balls were both green.

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3 votes

Answer:

The probability that Zuri draws a red ball is 2/5.

Given that Zuri draws a red ball, the probability that the discarded balls were both green is 66/171.

Explanation:

To find the probability that Zuri draws a red ball, we need to consider the total number of balls and the number of red balls in the box.

The total number of balls in the box is 8 (red) + 12 (green) = 20.

The probability of drawing a red ball is the number of red balls divided by the total number of balls: 8/20 = 2/5.

Now let's move on to the second part of the question. Given that Zuri draws a red ball, we want to find the probability that the discarded balls were both green.

When Zuri draws a red ball, there are 7 red balls and 12 green balls left in the box. The total number of balls remaining is 7 (red) + 12 (green) = 19.

To find the probability that both discarded balls were green, we need to consider the number of ways we can choose 2 green balls out of the remaining 12 green balls, divided by the total number of ways we can choose 2 balls from the remaining 19 balls.

The number of ways to choose 2 green balls out of 12 is given by the combination formula: C(12, 2) = 66.

The total number of ways to choose 2 balls out of 19 is given by the combination formula: C(19, 2) = 171.

Therefore, the probability that the discarded balls were both green, given that Zuri draws a red ball, is 66/171.

To summarize:

(i) The probability that Zuri draws a red ball is 2/5.

(ii) Given that Zuri draws a red ball, the probability that the discarded balls were both green is 66/171.

hope it helped <3

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