Math, asked by mischievousgirl58681, 3 months ago

Find the probability of drawing two red balls in succession from a bag containing 3 red and 6 black balls when (i) the ball is draw first is replaced (ii) it is not replaced

Answers

Answered by anuraggupta8938
0

Answer:

(i) 1/9

(ii) 1/12

Step-by-step explanation:

(i) probability of drawing 2 red balls uf the ball is draw first is replaced=3/9*3/9

=1/9

(ii) probability of drawing 2 red balls if it is not replaced =3/9*2/8

=1/12

Answered by dreamrob
0

Given,

The number of red balls = 3

The number of black balls = 6

To Find,

The probability of drawing two red balls in succession from a bag when the first ball is replaced =?

The probability of drawing two red balls in succession from a bag when the first ball is not replaced =?

Solution,

Total balls = number of red balls + number of black balls

Total balls = 3 + 6

Total balls = 9

The probability of drawing a red ball = The number of red balls / Total balls

The probability of drawing a red ball = 3 / 9

The probability of drawing a red ball = 1 / 3

The probability of drawing a second red ball if the ball is replaced = 3 / 9= 1/3

The probability of drawing two red balls in succession from a bag when the first ball is replaced = (1 / 3) * (1 / 3) = 1 / 9

If the ball is not replaced for second time,

The number of red balls = 3 - 1 = 2

Total balls = 9 - 1 = 8

The probability of drawing a second red ball if the ball is not replaced=2/8

The probability of drawing a second red ball if the ball is not replaced=1/4

The probability of drawing two red balls in succession from a bag when the first ball is not replaced = (1 / 3) * (1 / 4)

The probability of drawing two red balls in succession from a bag when the first ball is not replaced = 1 / 12

Hence, the probability of drawing two red balls in succession from a bag containing 3 red and 6 black balls when the first drawn ball is replaced is 1/9 and the probability of drawing two red balls in succession from a bag containing 3 red and 6 black balls when the first drawn ball is not replaced is 1/12.

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