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An urn contains balls numbered 1, 2, 3, First a ball is drawn from the urn, and then a fair coin is tossed the number of times as the number shown on the drawn ball. Find the expected number of heads

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  • $\begingroup$ @mathfan27543 trying to evaluate p.m.f $\endgroup$ – Rohit Pande Sep 5 '17 at 6:18
  • $\begingroup$ It would be helpful if you can show your work so far. $\endgroup$ – mathfan27543 Sep 5 '17 at 6:19
  • $\begingroup$ @mathfan27543 so firstly I kept no. Of heads as the random variable then I m trying to find the probability of the given random variable . Their I got stucked ~ $\endgroup$ – Rohit Pande Sep 5 '17 at 6:21
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I think I shouldn't completely answer this, but here is a hint: Be X the random variable you are interested in, Y a random variable with 2 states $(x_1,x_2)$ (This works for any countable number of states though). $$P(X\in B)=P(X\in B \cap \Omega)=P((X\in B)\cap(Y=x_1 \cup Y=x_2)) \\=P((X\in B)\cap(Y=x_1))+P((X\in B) \cap(Y=x_2)) \\= P(X\in B |Y=x_1)P(Y=x_1)+P(X\in B|Y=x_2)P(Y=x_2)$$

Feel free to ask if you don't understand a part of this.

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  • $\begingroup$ Hint2: what is the Probability to get 2 Heads, given that you have drawn the ball with number 1? $\endgroup$ – Felix B. Sep 5 '17 at 7:22
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Tree:

E(x) = Expected Number of Heads.

Pick 1 with a probability ($\frac{1}{3})$ - Toss the coin one time. E(X) = $\frac{1}{2}.\frac{1}{3}.1$

Pick 2 with a probability ($\frac{1}{3})$ - Toss two coins at a time E(X) = $\frac{1}{2}.\frac{1}{3}.1+\frac{1}{4}.\frac{1}{3}.2$

Pick 3 with a probability ($\frac{1}{3})$ - Toss three coins at a time E(X) =$ \frac{3}{8}.\frac{1}{3}.1+\frac{3}{8}.\frac{1}{3}.2+\frac{1}{8}.\frac{1}{3}.3$

Sum all, you get Expected Number of Heads = $\boxed{1}$

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