A bag contains 5 red, 4 blue and 3 green balls. A ball is taken out of the bag at random. Find the probability that the selected ball is (0 of red colour (ii) not green colour.

Solution

No. of red colour balls = 5
No. of blue colour balls = 4
No. of green colour balls = 3
Total no. of balls = 12
i.e., n( S) = 12
I. Let ‘A’ be the favourable outcomes of getting ‘of red colour’. Then,
i.e., n( A) = 5
Therefore,
$rightwards double arrow space space space space space straight P left parenthesis straight A right parenthesis space equals space fraction numerator straight n left parenthesis straight A right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 5 over 12$

II. Let 'B' be the favourable outcomes of getting not of green colour. Then
i.e., n( B) = 9
Therefore,

$rightwards double arrow space space space space space straight P left parenthesis straight B right parenthesis space equals space fraction numerator straight n left parenthesis straight B right parenthesis over denominator straight n left parenthesis straight S right parenthesis end fraction equals 9 over 12 equals 3 over 4.$

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Real Numbers

A bag contains 5 red, 4 blue and 3 green balls. A ball is taken out of the bag at random. Find the probability that the selected ball is (0 of red colour (ii) not green colour.

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