GATE CSE 2006 | Question: 21

Question

For each element in a set of size $2n$, an unbiased coin is tossed. The $2n$ coin tosses are independent. An element is chosen if the corresponding coin toss was a head. The probability that exactly $n$ elements are chosen is

• A. $\frac{^{2n}\mathrm{C}_n}{4^n}$
• B. $\frac{^{2n}\mathrm{C}_n}{2^n}$
• C. $\frac{1}{^{2n}\mathrm{C}_n}$
• D. $\frac{1}{2}$

Comments

As here they are talking about choosing n elementsout of 2n elements. so they can choose 1 element out of 2n or 2 elements out of 2n.....................2n elements out of 2n.

so total no of cases 2nc1 + 2nc2 + 2nc3+.............+2nc2n= 2^(2n)= 4^n

ways of getting n heads out of 2n tries = ²ⁿC_n

probability of getting exactly n heads and n tails = (1/2ⁿ)(1/2ⁿ)

number of ways = ²ⁿC_n/4ⁿ

Answer 1

Answer - A

Ways of getting $n$ heads out of $2n$ tries $= ^{2n}C_{n}.$

Probability of getting exactly $n$-heads and $n$-tails $= \left(\dfrac{1}{2^n}\right).\left(\dfrac{1}{2^n}\right)$

Number of ways $= \dfrac{^{2n}C_{n}}{4^n}.$

Comments

what if we do as below:
no. of ways to get n heads = 2nCn
no. of total outcome = 2n
therefore, probability of getting exactly n heads out of 2n tosses = 2nCn /2n

what does it calculate?

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How is it saying n heads out of 2n?

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We can simply solve this using Bionomial Distribution

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$\text{Binomial Distribution Formula :}  \;\; ^nC_{k} \; p^k \;(1-p)^{n-k}$

Probability that exactly $n$ elements are chosen out of $2n$ =

$^{2n}C_{n}\;\left(\dfrac{1}{2}\right)^n \left(1-\dfrac{1}{2}\right)^{2n-n}$

$^{2n}C_{n}\;\left(\dfrac{1}{2^n}\right)\left(\dfrac{1}{2^n}\right)$

$\dfrac{^{2n}C_{n}}{4^n}$

$Ans: A$

Answer 2

Answer :Option A 

Here is the link for theory- http://stattrek.com/probability-distributions/binomial.aspx
 

Comments

the best way to solve...

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This is easy to understand

Answer 3

Required Probability=$\frac{No. of Favourable Ways }{Total no. of Ways}$

No. of favourable ways= ²ⁿC_n  (bcoz select exactly n heads out of 2n tosses )

Total no. of ways= 2n tosses and each have 2 possibilities either H or T so total=$2^{2n}$=$4^n$ possibilities

So Ans is $\frac{\binom{2n}{n} }{4^{n}}$  which fits option A.

Comments

Finally found easiest answer for this problem thanks @rajesh

Answer 4

The question is mainly about probability of n heads out of 2n coin tosses.
P = 2nCn∗((1/2)^n)∗((1/2)^n) = (2nCn) / (4^n)