Question

A bag contains 6 white and 4 red balls. three balls are drawn at random. find the probability that one ball is red and the other two are white?

Answer 1

the answer will be 1/720 this is the probability














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Answer 2

↑ Here is your answer ↓
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Let's note down all the balls. 'W' is white and 'R' is red. Also number them.

$(W1) (W2) (W3) (W4) (W5) (W6) (R7)(R8)(R9)(R10)$

Now note down all the possibilities.

Possiblities with 'W1' =

$(123) (124) (125) (126) (127) (128) (129) (1210)$

$(132) (134) (135) (136) (137) (138) (139) (1310)$

... this can be possible until (1910)

Similarly, note down with remaining 10 numbers

Total number of possibilities:

There are '8' possibilities with each number,

Each number can be crossed with '9' other numbers,

You should repeat this with all other '10 numbers.

8 * 9 * 10 = 720

One is red and other two are white: 

2 White and 1 Red from starting with 'W1' = 4

Until W6, there will be 4 so = 4 * 6 = 24

2 White and 1 Red from starting with 'R7' = 6

Until R10, there will be 6 so = 6 * 4 = 24

24 + 24 = 48

Final answer:

$48/720 = 0.067$

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Glad to help you.
@EmadAhamed