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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Three cards are drawn successively with replacement from a well shuffled deck of 52 playing cards. If getting a card of spade is considered a success, find the probability distribution of the number of success.

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  • Once we write down the sample space for the problem we can find the favorable outcomes in each case and find their probabilities and arrange as a probability distribution.
Step 1:
Let $X$ denote the no of spade cards in three draws.
$\therefore x$ can take values 0,1,2,3
$P(X=0)$=Probability of not getting a spade card.
Probability of getting a spade card is $P(S)=\large\frac{13}{52}$
$\Rightarrow \large\frac{1}{4}$
Probability of not getting a spade card is $P(F)=1-\large\frac{1}{4}$
$\Rightarrow \large\frac{3}{4}$
Step 2:
$P(X=0)=P(FFF)$
$\qquad\quad\;\;=\large\frac{3}{4}\times\large\frac{3}{4}\times\large\frac{3}{4}$
$\qquad\quad\;\;=\large\frac{27}{64}$
$P(X=1)=P(SFF)+P(FSF)+P(FFS)$
$\qquad\quad\;\;=\large\frac{1}{4}\times\large\frac{3}{4}\times\large\frac{3}{4}+\large\frac{3}{4}\times\large\frac{1}{4}\times\large\frac{3}{4}+\large\frac{3}{4}\times\large\frac{3}{4}\times\large\frac{1}{4}$
$\qquad\quad\;\;=3\times\large\frac{3^2}{4^3}$
$\qquad\quad\;\;=\large\frac{27}{64}$
$P(X=2)=P(SSF)+P(FSS)+P(SFS)$
$\qquad\quad\;\;=\large\frac{1}{4}\times\large\frac{1}{4}\times\large\frac{3}{4}+\large\frac{3}{4}\times\large\frac{1}{4}\times\large\frac{1}{4}+\large\frac{1}{4}\times\large\frac{3}{4}\times\large\frac{1}{4}$
$\qquad\quad\;\;=3\times\large\frac{3^3}{4^3}$
$\qquad\quad\;\;=\large\frac{9}{64}$
$P(X=3)=P(SSS)$
$\qquad\quad\;\;=\large\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}$
$\qquad\quad\;\;=\large\frac{1}{64}$
Step 3:
Hence the required probability is
answered Oct 2, 2013 by sreemathi.v
 

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