A biased coin with probabilty p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $\frac{2}{5},$ then p equals:
A$\frac{1}{3}$
B$\frac{2}{3}$
C$\frac{2}{5}$
D$\frac{3}{5}$
Download our app for free and get started
A$\frac{1}{3}$
p is the probability of getting head.
q = 1 - p is the probability of getting tail.
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
If A and B are two events associated to a random experiment such that $\text{P}(\text{A}\cap\text{B})=\frac{7}{10}$ and $\text{P(B)}=\frac{17}{20}$, then P(A|B) =
Mark the correct alternative in the following question:
The probability that a person is not a swimmer is 0.3. The probability that out of 5 persons 4 are swimmers is:
Mark the correct alternative in the following question:
Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If $\frac{\text{P(X = r})}{\text{P(X = n} -\text{r})}$ is independent of n and r, then p equals:
If A and B are two events such that P(A) = 0.4, P(B) = 0.3 and $\text{P}(\text{A}\cup\text{B})=0.5,$ then $\text{P}(\overline{\text{B}}\cap\text{A})$ equals.