Let C_1 and C_2 be two biased coins such that | vedclass

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

Question

(B) For $C_1$,$P(H) = \frac{2}{3}$. The number of heads $\alpha$ follows a binomial distribution $B(2, \frac{2}{3})$.$P(\alpha = 0) = (\frac{1}{3})^2

Vedclass · Accepted Answer

$\frac{20}{81}$

Vedclass · Answer

(B) For $C_1$,$P(H) = \frac{2}{3}$. The number of heads $\alpha$ follows a binomial distribution $B(2, \frac{2}{3})$.$P(\alpha = 0) = (\frac{1}{3})^2 = \frac{1}{9}$$P(\alpha = 1) = 2 	imes \frac{2}{3} 	imes \frac{1}{3} = \frac{4}{9}$$P(\alpha = 2) = (\frac{2}{3})^2 = \frac{4}{9}$For $C_2$,$P(H) = \frac{1}{3}$. The number of heads $\beta$ follows a binomial distribution $B(2, \frac{1}{3})$.$P(\beta = 0) = (\frac{2}{3})^2 = \frac{4}{9}$$P(\beta = 1) = 2 	imes \frac{1}{3} 	imes \frac{2}{3} = \frac{4}{9}$$P(\beta = 2) = (\frac{1}{3})^2 = \frac{1}{9}$The roots of $x^2 - \alpha x + \beta = 0$ are real and equal if the discriminant $D = \alpha^2 - 4\beta = 0$,i.e.,$\alpha^2 = 4\beta$.Possible pairs $(\alpha, \beta)$ satisfying this are $(0, 0)$ and $(2, 1)$.Probability $= P(\alpha=0)P(\beta=0) + P(\alpha=2)P(\beta=1)$$= (\frac{1}{9} 	imes \frac{4}{9}) + (\frac{4}{9} 	imes \frac{4}{9}) = \frac{4}{81} + \frac{16}{81} = \frac{20}{81}$.

Similar Questions

$A$ binomial random variable $X$ satisfies $9 \cdot P(X=4) = P(X=2)$ when $n=6$. Then $p$ is equal to

The probability of guessing correctly at least $7$ out of $10$ answers in a "True" or "False" test is = $ . . . . . . $

$A$ die is thrown $6$ times. If 'getting an odd number' is a success,what is the probability of $5$ successes?

$A$ multiple-choice examination consists of $5$ questions. Each question has $3$ alternative answers,of which only $1$ is correct. What is the probability that a student will get $4$ or more correct answers?

The probability that a person who undergoes a certain operation will survive is $0.2$. If $5$ patients undergo similar operations,then the probability that exactly four will survive is

Vedclass Test Series

Exam Paper Generator

Online Exam Module