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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Two balls are drawn from an urn containing 2 white, 3 red and 4 black balls one by one without replacement. What is the probability that at least one ball is red?

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Toolbox:
  • \(P(A\cap\;B)=P(A)\;P(B)\)
  • P (A $\cap$ B) = P(A) + P(B) - P(A $\cup$ B)
  • \(\;P(B/A)=\large \frac{P(A\cap\:B)}{P(A)}\)\(\:;P(A/B)=\large \frac{P(A\cap\;B)}{P(B)}\)
Step 1:
Let $A$ be the event of not getting a red ball in first draw and $B$ be the event of not getting a red ball in second draw,
Then the required probability=Probability that at least one ball should be red.
$\Rightarrow$ 1-Probability that none is red.
$\Rightarrow 1-P(A\; and\; B)$
$\Rightarrow 1-P(A\cap B)$
$\Rightarrow 1-P(A)P( B/A)$
Step 2:
Let $P(A)$ =Probability of not getting a red ball in first draw.
(i.e)$P(A)$=Probability of getting another colour(white or black) ball in first draw.
$\Rightarrow P(A)=\large\frac{6}{9}=\frac{2}{3}$
Now that are 5 other color balls (white or black) and 3 red balls.
The other color ball can be drawn in $5C_1$ ways
$\therefore P(B/A)=\large\frac{5}{8}$
Step 3:
Hence the required probability is $1-P(A)P(B/A)$
$\Rightarrow 1-\large\frac{2}{3}\times \frac{5}{8}$
$\Rightarrow \large\frac{7}{12}$
answered Sep 30, 2013 by sreemathi.v
 

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