A bag contains 3 white, 3 black and 2 red balls. One by one three balls are drawn without replacing

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14.Probability

hard

A bag contains $3$ white, $3$ black and $2$ red balls. One by one three balls are drawn without replacing them. The probability that the third ball is red, is

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{1}{4}$

Solution

(d) Let $R$ stand for drawing red ball $B$ for drawing black ball and $W$ for drawing white ball.

Then required probability

$ = P(WWR) + P(BBR) + P(WBR) + P(BWR) + P(WRR) + $

$P(BRR) + P(RWR) + P(RBR).$

$ = \frac{{3.2.2}}{{8.7.6}} + \frac{{3.2.2}}{{8.7.6}} + \frac{{3.3.2}}{{8.7.6}} + \frac{{3.3.2}}{{8.7.6}} + \frac{{3.2.1}}{{8.7.6}}$

$ + \frac{{3.2.1}}{{8.7.6}} + \frac{{2.3.1}}{{8.7.6}} + \frac{{2.3.1}}{{8.7.6}}$

$ = \frac{2}{{56}} + \frac{2}{{56}} + \frac{3}{{56}} + \frac{3}{{56}} + \frac{1}{{56}} + \frac{1}{{56}} + \frac{1}{{56}} + \frac{1}{{56}} = \frac{1}{4}$.

Standard 11

Mathematics

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$A:$ getting an even number on the first die.

$B:$ getting an odd number on the first die.

$C:$ getting the sum of the numbers on the dice $\leq 5$

State true or false $:$ (give reason for your answer)

Statement : $A=B^{\prime}$

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normal

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(AIEEE-2003)

A bag contains $3$ white, $3$ black and $2$ red balls. One by one three balls are drawn without replacing them. The probability that the third ball is red, is

Solution

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Two dice are thrown. The events $A,\, B$ and $C$ are as follows:

$A:$ getting an even number on the first die.

$B:$ getting an odd number on the first die.

$C:$ getting the sum of the numbers on the dice $\leq 5$

State true or false $:$ (give reason for your answer)

Statement : $A=B^{\prime}$