A fair coin is tossed 100 times. The chance o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $n = 100$ be the number of tosses and $p = q = \frac{1}{2}$ be the probability of getting a head or tail in a single toss.The probability of g

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Let $n = 100$ be the number of tosses and $p = q = \frac{1}{2}$ be the probability of getting a head or tail in a single toss.The probability of getting exactly $r$ heads is given by the binomial distribution: $P(X = r) = \binom{n}{r} p^r q^{n-r} = \binom{100}{r} (\frac{1}{2})^{100}$.We want the probability of getting an even number of heads,which is $S = \sum_{r 	ext{ is even}} \binom{100}{r} (\frac{1}{2})^{100}$.Using the binomial expansion,we know $(p+q)^n = \sum_{r=0}^{n} \binom{n}{r} p^r q^{n-r} = 1$ and $(q-p)^n = \sum_{r=0}^{n} \binom{n}{r} (-p)^r q^{n-r}$.For $p=q=\frac{1}{2}$,$(q-p)^n = 0^n = 0$ for $n \ge 1$.Thus,$\sum_{r=0}^{n} \binom{n}{r} (\frac{1}{2})^n = 1$ and $\sum_{r=0}^{n} \binom{n}{r} (-1)^r (\frac{1}{2})^n = 0$.Adding these two equations: $2 \sum_{r 	ext{ is even}} \binom{n}{r} (\frac{1}{2})^n = 1 + 0 = 1$.Therefore,the probability of an even number of heads is $S = \frac{1}{2}$.

$A$ fair coin is tossed $100$ times. The chance of getting a head an even number of times is

Similar Questions

$A$ fair coin is tossed a fixed number of times. If the probability of getting $5$ tails is same as the probability of getting $7$ tails,then the probability of getting $3$ tails is

If $X$ has a binomial distribution with mean $np$ and variance $npq$,then $\frac{P(X = k)}{P(X = k - 1)}$ is

$A$ coin is tossed $n$ times. If the probability of getting $6$ heads is equal to the probability of getting $8$ heads,then find the value of $n$.

The mean and variance of a binomial variable $X$ are $2$ and $1$ respectively. Then,the probability that $X$ takes values greater than $1$ is

The mean and variance of a binomial variate $X$ are $16/5$ and $48/25$ respectively. If $P(X > 1) = 1 - K (3/5)^7$, then $5 K =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module