Applied Mathematics

Probability Distribution

Easy

The variance of the number of heads in two tosses of a coin is :

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{8}$

✅ Correct Option: 1

Let X be the number of heads in two tosses. X follows binomial with $n = 2$ , $p = q = \frac{1}{2}$ . Variance $= npq = 2 \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2}$ .

2023: 22 May Shift 3

If a random variable X follows binomial distribution with mean 5 and variance $\frac{5}{2}$ , then $P(X \leq 9)$ is :

2026: 15 May Shift 1

A coin is tossed until a head appears, or the tail appears 4 times in succession, then the probability distribution of the number of tosses is

2023: 23 May Shift 3

A shopkeeper wants to check the average number of cars sold per call. Past record of sales is shown below:

Sale of cars (Units)	0	1	2	3
Probabilities	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$\frac{1}{30}$

The expected number of cars sold is :

2026: 12 May Shift 2

Ajay is a vendor who informs that there are 2% rotten apples in the fruit bag. Find the probability that a sample of 10 apples will include not more than one rotten apple.

2023: 22 May Shift 3

If the probability distribution of X is :

X	2	3	4	5	6
P(X)	1/15	2/15	3/15	4/15	5/15

Then variance is equal to :

2023: 11 Aug Shift 1

A random variable X has a probability distribution P(X) of the following form, where k is some unknown constant:

P(X = 0) = k

P(X = 1) = 2k

P(X = 2) = 3k

P(X = other values) = 0

Then, find the value of 1/k.

The variance of the number of heads in two tosses of a coin is :

Solution

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