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

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Ans

Answer 1

 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)
Answer: (A)

Answer 2

Option A

 

Answer 3

we need exactly $n$ Heads, so assume them as string of heads and tails(think like string of $0$ & $1$)

$\underbrace{H \\\  H \\\ H .. . .. .H}_{n-times} \\\ \underbrace{T \\\ T \\\ T......T}_{n-times}$

total permutation of strings for exactly $n$ heads :  $\frac{2n!}{n! n!}$

total permutation of string of heads and tails of length $2n$  :  $(2)^{2n} = 4^n$

$$\frac{\frac{2n!}{n! n!}}{4^n}$$

Answer 4

Ans