36 votes 36 votes 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 $\frac{^{2n}\mathrm{C}_n}{4^n}$ $\frac{^{2n}\mathrm{C}_n}{2^n}$ $\frac{1}{^{2n}\mathrm{C}_n}$ $\frac{1}{2}$ Probability gatecse-2006 probability binomial-distribution normal + – Rucha Shelke asked Sep 17, 2014 Rucha Shelke 8.1k views answer comment Share Follow See 1 comment See all 1 1 comment reply Dileep kumar M 6 commented Nov 23, 2017 reply Follow Share 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 = 2nCn probability of getting exactly n heads and n tails = (1/2n)(1/2n) number of ways = 2nCn/4n 7 votes 7 votes Please log in or register to add a comment.
2 votes 2 votes We can also put small values for n and eliminate options. Jhaiyam answered Jul 20, 2020 Jhaiyam comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes sample space : 2 * 2 * 2 … *2 ( 2n) times i.e. $2^{2n}$ which is $4^{n}$ favourable outcomes : chosing exactly n heads out of 2n tosses which is $^{2n}C_{n}$ (option A) $\frac{^{2n}C_{n}}{4^{n}}$ shashankrustagi answered Dec 19, 2020 shashankrustagi comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes 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) varunrajarathnam answered Jun 2, 2021 varunrajarathnam comment Share Follow See all 0 reply Please log in or register to add a comment.