The probability of getting 4 heads in 8 throw | vedclass

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

Question

(D) The probability of getting $k$ successes in $n$ independent Bernoulli trials is given by the binomial distribution formula: $P(X = k) = {}^nC_k \c

Vedclass · Accepted Answer

$\frac{^8C_4}{2^8}$

Vedclass · Answer

(D) The probability of getting $k$ successes in $n$ independent Bernoulli trials is given by the binomial distribution formula: $P(X = k) = {}^nC_k \cdot p^k \cdot q^{n-k}$.Here,$n = 8$ (total number of throws),$k = 4$ (number of heads required),$p = \frac{1}{2}$ (probability of getting a head),and $q = 1 - p = \frac{1}{2}$ (probability of getting a tail).Substituting these values into the formula:$P(X = 4) = {}^8C_4 \cdot (\frac{1}{2})^4 \cdot (\frac{1}{2})^{8-4}$$P(X = 4) = {}^8C_4 \cdot (\frac{1}{2})^4 \cdot (\frac{1}{2})^4$$P(X = 4) = {}^8C_4 \cdot (\frac{1}{2})^8$$P(X = 4) = \frac{{}^8C_4}{2^8}$.Thus,the correct option is $(d)$.

The probability of getting $4$ heads in $8$ throws of a coin is:

Similar Questions

Two dice are thrown. The probability that the sum of the points on two dice will be $7$ is:

$2$ dice are thrown. Find the odds against getting the sum $6$.

There are three events $A, B, C$ one of which must and only one can happen. The odds are $8$ to $3$ against $A$ and $5$ to $2$ against $B$. Find the odds against $C$.

The letters of the word $SOCIETY$ are placed in a row. What is the probability that the three vowels come together?

Let $0 < P(A) < 1$,$0 < P(B) < 1$ and $P(A \cup B) = P(A) + P(B) - P(A)P(B).$ Then

Vedclass Test Series

Exam Paper Generator

Online Exam Module