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Bunuel
A bag contains 4 blue, 5 white and 6 green balls. Two balls are drawn at random without replacement . What is the probability that one ball is white?

A. 2/35
B. 10/21
C. 1/2
D. 3/4
E. 4/5

Given: A bag contains 4 blue, 5 white and 6 green balls. Two balls are drawn at random without replacement .
Asked: What is the probability that one ball is white?

Total balls = 15
Total ways to draw 2 balls out of 15 = \(^{15}C_2 = 105\)

Number of ways so that one balls is white = \(^5C_1 * ^{10}C_1 = 5 * 10 = 50\)

Probability =\( \frac{50 }{105} = \frac{10}{21}\)

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hi, is this approach right?

first white = 5/15
second any other color = 10/14
total prob: 5/15*10/14*2
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hi, is this approach right?

first white = 5/15
second any other color = 10/14
total prob: 5/15*10/14*2

The answer will be revealed on Saturday but
yes, it's a correct approach and answer
.
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White ball can be drawn Frist time or 2nd time:
Probability if 1st ball is white : 5/15 * 10/14 (10/14 is red + blue balls / balance no of balls in bag as its w/o replacement) ---------------------- 1
Probability if 2nd ball is white : 10/15 *5/14 ---------------- 2
Adding 1 and 2
10/42 +10/42 = 20/42 => 10/21


IMO B
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