GATE CSE
First time here? Checkout the FAQ!
x
+3 votes
598 views

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

  1. $\frac{^{2n}\mathrm{C}_n}{4^n}$
  2. $\frac{^{2n}\mathrm{C}_n}{2^n}$
  3. $\frac{1}{^{2n}\mathrm{C}_n}$
  4. $\frac{1}{2}$
asked in Probability by Loyal (4k points)   | 598 views

3 Answers

+9 votes
Best answer

answer - A

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

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

number of ways = 2nCn/4n

answered by Boss (9.1k points)  
selected by
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?
How is it saying n heads out of 2n?
+5 votes

Answer :Option A 

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

answered by Active (2k points)  
+3 votes

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

No. of favourable ways= 2nC (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.

answered by Veteran (17k points)  
edited by


Top Users Jul 2017
  1. Bikram

    3946 Points

  2. manu00x

    2464 Points

  3. Debashish Deka

    1842 Points

  4. joshi_nitish

    1650 Points

  5. Arjun

    1268 Points

  6. Hemant Parihar

    1184 Points

  7. Arnab Bhadra

    1100 Points

  8. Shubhanshu

    1052 Points

  9. Ahwan

    900 Points

  10. rahul sharma 5

    692 Points


24,016 questions
30,946 answers
70,303 comments
29,333 users