Question

An urn contains 9 red, 7 white and 4 black balls. If two balls are drawn at random find the probability that

(i) both the balls are red.

(ii) one is white and other is red.

(iii) the balls are of same Colour.

Or

A box contains 10 bulbs, of which just three are defective. If at random a sample of five bulbs is drawn, find the probabilities that the sample contains

(i) exactly one defective bulb.

(ii) exactly two defective bulbs.

(iii) no defective bulbs.

Answer 1

There are 20 balls in the bag out of 2 balls can be drawn in ${}^{20}{C}_2$ways.

So, total number of out comes =${}^{20}{C}_2=190.$

(i) There are 9 red balls of which 2 balls can be drawn in ${}^9{C}_2$ways.

$\therefore$ Favourable number of outcomes =${}^9{C}_2=36$

So, required probability $=\frac{36}{190}=\frac{18}{95}$

(ii) Out of 7 white balls, one white ball can be drawn in ${}^7{C}_1$ways and out of 9 red balls. one red ball can be drawn in ${}^9{C}_1$ways.

$\therefore$ Favourable number of outcomes =${}^7{C}_1{\times }^9{C}_1=63$

So, required probability =$\frac{63}{190}$

(iii) Two balls are drawn of same colour mean that either both are red or both are white or both are black, out of 9 red balls two red balls are drawn in ${}^9{C}_2$ways. Similarly two white balls can be drawn from 7 white balls, in ${}^7{C}_2$ways and two black balls drawn from 4 black balls in ${}^4{C}_2$ ways.

$\therefore$ The number of ways of drawing 2 balls of same colour

${=}^9{C}_2{+}^7{C}_2{+}^4{C}_2=36+21+6=63$

So, required probability =$\frac{63}{190}$

Or

Out of 10 bulbs 5 can be drawn in ${}^{10}{C}_5$ways.

So, total number of outcomes

=${}^{10}{C}_5=\frac{10!}{5!5!}=252$

(i) There are 3 defective and 7 non-defective bulbs.The number of ways of selecting one defective bulb out of 3 and 4 non-defectvie bulbs out of 7 is ${}^3{C}_1{\times }^7{C}_4$.

$\therefore$ Favourable number of outcomes ${}^3{C}_1{\times }^7{C}_4$

$=\frac{3!}{1!2!}\times \frac{7!}{3!4!}=3\times 35=105$

So, required probability $=\frac{105}{252}=\frac{5}{12}$

(ii) The number of ways of selecting 2 defective bulbs out of 3 and 3 non-defective bulbs out of 7 is ${}^3{C}_2{\times }^7{C}_3$.

$\therefore$ Favourable number of outcomes =${}^3{C}_2{\times }^7{C}_3$

=$=3\times 35=105$

So, required probability =$\frac{105}{252}=\frac{5}{12}$

(iii) No defective bulb means all non-defective bulbs. The number of ways of selecting all 5 non-defective bulbs out of 7 is ${}^7{C}_5$

$\therefore$ Favourable number of outcomes =${}^7{C}_5=21$

So, required probability =$\frac{21}{252}=\frac{1}{12}$