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In a box, there are $2$ red, $3$ black and $4$ blue coloured balls. The probability of drawing $2$ blue balls in sequence without replacing, and then drawing $1$ black ball from this box is _________ %.
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Where I distinguished?

I have taken any one
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Thugh I understand the ans

thanks
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2 Answers

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@srestha 

Same thing i had done also when i was getting 1/14 i just cancel out for easier calculations 

1/14 = 0.07142 = 7.14% only

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but where is my fault?
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U have not considered the number of ways we can pick up blue balls from those 9 balls,

we can pick the blue balls in (4C2 / 9C2) ways, then we are choosing the black ball, so then multiply it by 3/7.. :)

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In the box we have --> Red--> 2 Black--> 3 Blue--> 4

Now we are picking up 2 blue balls at random which can be done in (4C2 / 9C2) ways

after that we are picking up a black ball, number of ways to do that --> (3C1 / 7C1) ways

so we are having probability --> (4C2 / 9C2) x (3C1 / 7C1)

which gives us-->  0.07142

As the ans is asked in % so, --> 0.07142 * 100 = 7.14 %                                                  

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yes , but @akash.dinkar12 done rightly, as it picked one by one :)

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