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Bunuel
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Neodymium
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red = 5
white= 4
total =9

no replacement allowed
draw 3 balls

find: P (at least 2 red)=

2 case here.

case- 1 2 red, 1 white.

for this, 3!/2! = 3 ways.

since no replacement, RRW
5/9 * 4/8 * 4/7. and there are 3 ways.
so, 3{5/9* 4/8 * 4/7}

now case 2, all 3 red/ for that only one way.
so, 5/9 * 4/8 * 3/7

add both cases.

we get 25/42

option D
Bunuel
A bag contain 5 red balls and 4 white balls. Three balls are withdrawn without replacement. What is the probability of drawing at least 2 red balls?

A. 5/42
B. 5/18
C. 17/42
D. 25/42
E. 13/18
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vasu1104
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red = 5
white= 4
total =9

no replacement allowed
draw 3 balls

find: P (at least 2 red)=

2 case here.

case- 1 2 red, 1 white.

for this, 3!/2! = 3 ways.

since no replacement, RRW
5/9 * 4/8 * 4/7. and there are 3 ways.
so, 3{5/9* 4/8 * 4/7}

now case 2, all 3 red/ for that only one way.
so, 5/9 * 4/8 * 3/7

add both cases.

we get 25/42

option D
Bunuel
A bag contain 5 red balls and 4 white balls. Three balls are withdrawn without replacement. What is the probability of drawing at least 2 red balls?

A. 5/42
B. 5/18
C. 17/42
D. 25/42
E. 13/18
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