GATE Mechanical Engineering (ME) 2018 Shift 1 Solved Paper

Question : 24 of 65

Marks: +1, -0

Four red balls, four green balls and four blue balls are put in a box. Three balls are pulled out of the box at random one after another without replacement. The probability that all the three balls are red is

$1 / 72$
$1 / 55$
$1 / 36$
$1/27$

Validate

Solution:

Probability that all the three balls are red is

{P}= {R} . {R} . {R}

P=\;{4}/{12} × \;{3}/{11} × \;{2}/{10}=\;{24}/{1320}=\;{1}/{55}

Alter:
The required probability can also be obtained as,
Probability

= \;{4 C_{3}}/{12 C_{3}}=\;{4 × 3 × 2}/{3 × 2 × 1} × \;{3 × 2 × 1}/{12 × 11 × 10}=\;{1}/{55}

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