Question

ii) A bag contains 4 green and 6 red
balls. 2 balls are drawn at random
one by one without replacement.
What is the chance a green ball is
drawn each time?
​

Answer 1

Answer:

Initially, there are 6 red balls and 4 green balls.

P(R1) = Probability of getting a red ball on the first draw = (6 / 10).

If the first draw yields a red ball; there are 5 red balls and 4 green balls left.

P(R2) = Probability of getting a red ball on the first draw = (5 / 9).

Therefore, probability of getting two red balls from the first two draws

= P(R1)*P(R2) = (6 / 10)*(5 / 9) = (1 / 3) = 0.3333.

Explanation:

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