Algebra

Probability

Hard

applied

The sum and the product of the mean and variance of binomial distribution are 1.8 and 0.8 respectively. The probability of atleast one success is:

$\left(\frac{4}{5}\right)^5$

$\left(\frac{4}{5}\right)^4$

$1-\left(\frac{4}{5}\right)^5$

$1-\left(\frac{4}{5}\right)^4$

✅ Correct Option: 3

With mean $m=np$ and variance $v=npq$ , $m+v=1.8$ and $mv=0.8$ , so $m$ and $v$ are the roots of $t^2-1.8t+0.8=0$ , namely $1$ and $0.8$ .

The mean exceeds the variance, so $np=1$ , $q=0.8$ , $p=0.2$ and $n=5$ .

Hence $P(X\ge1)=1-\left(\frac{4}{5}\right)^5$ .

2022: 4 Aug Shift 1

If A and B are two independent events with $P(A) = \frac{1}{5}$ and $P(B) = \frac{1}{3}$ , then $P(A'/B)$ is:

2022: 23 Aug Shift 1

The probability that a student knows the answer, is:

2025: 3 June Shift 2

The curve $y = f(x)$ is normal probability curve, then which of the following statements are correct?

(A) mean, median and mode of the distribution coincide.

(B) the area bounded by the curve $y = f(x)$ and $x$ -axis is one unit.

(D) $y$ -axis is an asymptote to the curve.

Choose the correct answer from the options given below:

2025: 29 May Shift 2

Let A and B be two events. Then which of the following statements are TRUE?

(A) $P(B|A) = \frac{P(A \cap B)}{P(A)}$ , provided $P(A) \neq 0$

(B) $P(B') = 1 + P(B)$

(D) $P(A \cap B) = P(A).P(B)$ If A and B are independent events

Choose the correct answer from the options given below:

2026: 29 May Shift 2

If A, B and C are three independent events such that P(A)=P(B)=P(C)=x, then P(at least two of A, B, C occur) =

2026: 22 May Shift 1

Let E and F are two independent events such that $P(E)=0.35$ and $P(E\cup F)=0.60$ then which of following statements are TRUE?

(A) $P(F)=\frac{5}{13}$

(B) $P(E|\overline{F})=0.35$

(D) $P(\overline{E}|\overline{F})=0.65$

Choose the correct answer from the options given below:

Solution