Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

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?

Can you answer this question?

1 Answer

0 votes
  • \(P(A\cap\;B)=P(A)\;P(B)\)
  • 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App