Math, asked by ishikapatidar1251, 9 months ago

A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:

A) 123/897 B) 23/67 C) 7/44 D) 12/45

Answers

Answered by omarkhalily75
0

Answer:

7/44

Step-by-step explanation:

Answer: C) 7/44

Explanation:

Here, n(E) = 7C1×5C1×5C1

And, n(S) = 12C1*11C1*10C1

P(S) = 7*6*512*11*10 = 7/44

Answered by dreamrob
0

Given,

The total green balls = 7

The total black balls = 5

To Find,

The probability of all three balls being green, if the balls are not replaced=?

Solution,

Total balls in the bag = 5 + 7 = 12

The balls are not replaced which means each time the number of balls reduces by 1.

The total ways = 12C1*11C1*10C1

The total ways = 12 * 11 * 10

The total ways = 1320

The total ways of choosing 3 green balls = 7C1*6C1*5C1

The total ways of choosing 3 green balls = 7 * 6 * 5

The total ways of choosing 3 green balls = 210

Probability = 210 / 1320

Probability = 7 / 44

Hence, the probability of all three balls being green, if the balls are not replaced is 7/44. Option(c) is the correct answer.

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