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Q.A lady's bag contains 2 black and 1 red pens. One pen is drawn at random and then put back in the box after noting its colour. The process is repeated again. If X denotes the number of red pens recorded in the two draws. Describe X.

Chandigarh CbseNCERTSubjective· 3mImportance★★★★★
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With replacement each draw is independent with P(red)=13P(\text{red})=\tfrac13; XX (reds in 2 draws) is Binomial B(2,13)B(2,\tfrac13) with distribution 49,49,19\tfrac49,\tfrac49,\tfrac19.

Binomial: P(X=r)=(nr)pr(1−p)n−rP(X=r)=\binom{n}{r}p^{r}(1-p)^{n-r}, here n=2n=2, p=P(red)=13p=P(\text{red})=\dfrac13, q=23q=\dfrac23. Mean E(X)=npE(X)=np, variance =npq=npq.

  1. Setup. The bag has 2 black and 1 red pen; drawing with replacement keeps p=13p=\dfrac13 constant over the n=2n=2 independent draws. So X∈{0,1,2}X\in\{0,1,2\}.
  2. Probabilities.
  • P(X=0)=(20)(13)0(23)2=49P(X=0)=\binom20\left(\tfrac13\right)^0\left(\tfrac23\right)^2=\dfrac49
  • P(X=1)=(21)(13)1(23)1=2⋅13⋅23=49P(X=1)=\binom21\left(\tfrac13\right)^1\left(\tfrac23\right)^1=2\cdot\dfrac13\cdot\dfrac23=\dfrac49
  • P(X=2)=(22)(13)2(23)0=19P(X=2)=\binom22\left(\tfrac13\right)^2\left(\tfrac23\right)^0=\dfrac19
  1. Distribution table.
XX012
P(X)P(X)49\tfrac4949\tfrac4919\tfrac19

Check: 49+49+19=1.\dfrac49+\dfrac49+\dfrac19=1.

4. Mean & variance. E(X)=np=2×13=23E(X)=np=2\times\dfrac13=\dfrac23; Var⁡(X)=npq=2×13×23=49.\operatorname{Var}(X)=npq=2\times\dfrac13\times\dfrac23=\dfrac49.

✓Final answer

X∼B ⁣(2,13)X\sim B\!\left(2,\tfrac13\right) with P(0)=49, P(1)=49, P(2)=19P(0)=\tfrac49,\ P(1)=\tfrac49,\ P(2)=\tfrac19; E(X)=23E(X)=\tfrac23, Var⁡(X)=49\operatorname{Var}(X)=\tfrac49.

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