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  • A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that (ii) atleast one will be green?

1.(ii)   A box contains \small 10 red marbles, \small 20  blue marbles and \small 30 green marbles. \small 5 marbles are drawn from the box, what is the probability that (ii) atleast one will be green? 

Answers (1)

best_answer

Given, 

No. of red marbles = 10

No. of blue marbles = 20

No. of green marbles = 30

Total number of marbles = 10 + 20 + 30 = 60

Number of ways of drawing 5 marbles from 60 marbles = ^{60}\textrm{C}_{5}

(ii). We know,

The probability that at least one marble is green = 1 - Probability that no marble is green

Now, Number of ways of drawing no green marbles = Number of ways of drawing only red and blue marbles 

= ^{(20+10)}\textrm{C}_{5} = ^{30}\textrm{C}_{5}

\therefore The probability that no marble is green = \frac{^{30}\textrm{C}_{5}}{^{60}\textrm{C}_{5}}

\therefore The probability that at least one marble is green = 1-\frac{^{30}\textrm{C}_{5}}{^{60}\textrm{C}_{5}}

Posted by

HARSH KANKARIA

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