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 Option 1: X, 1, 2, 3, 4, P(x), 1/2, 1/4, 1/8, 1/8 Option 2: X, 1, 2, 3, 4, P(x), 1/2, 1/4, 1/4, 1/8 Option 3: X, 1, 2, 3, 4, P(x), 1/2, 1/8, 1/4, 1/4 Option 4: X, 1, 2, 3, 4, P(x), 1/2, 1/8, 1/8, 1/4

Answer: X, 1, 2, 3, 4, P(x), 1/2, 1/4, 1/8, 1/8. Solution: Tossing stops at the first head or after four tails. So P(1)=P(H)=1/2, P(2)=P(TH)=1/4, P(3)=P(TTH)=1/8, and P(4)=P(TTTH)+P(TTTT)=1/16+1/16=1/8. These sum to 1.

Applied Mathematics

Probability Distribution

Medium

applied

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

X	1	2	3	4
P(x)	1/2	1/4	1/8	1/8

X	1	2	3	4
P(x)	1/2	1/4	1/4	1/8

X	1	2	3	4
P(x)	1/2	1/8	1/4	1/4

X	1	2	3	4
P(x)	1/2	1/8	1/8	1/4

✅ Correct Option: 1

Tossing stops at the first head or after four tails. So $P(1)=P(H)=\frac{1}{2}$ , $P(2)=P(TH)=\frac{1}{4}$ , $P(3)=P(TTH)=\frac{1}{8}$ , and $P(4)=P(TTTH)+P(TTTT)=\frac{1}{16}+\frac{1}{16}=\frac{1}{8}$ . These sum to $1$ .

2026: 21 May Shift 2

The random variable X has a probability distribution $P(X)$ of the following form, where $k$ is some real number.

$P(X=x) = \begin{cases}k, & \text{if } x=0\\\dfrac{k}{2}, & \text{if } x=1 \text{ or } 2\\0, & \text{otherwise}\end{cases}$

On the basis of the above information, match List-I with List-II

LIST-I		LIST-II
A. Value of $k$		I. $\dfrac{1}{4}$
B. $P(X<2)$		II. 0
C. $P(X>1)$		III. $\dfrac{1}{2}$
D. $P(X>2)$		IV. $\dfrac{3}{4}$

Choose the correct answer from the options given below:

2023: 11 Aug Shift 1

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

2022: 25 May Shift 1

If the probability distribution of a random variable X is given as

$x_i$	0	1	2	3
$p_i$	$2k^2$	$k^2$	$3k^2$	$k$

Then the mean of X is

2026: 18 May Shift 2

A pair of dice is thrown 5 times. If getting a total of 7 is considered a success, then

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. Probability of no success	I. $26\left(\dfrac{1}{6}\right)^5$
B. Probability of 5 successes	II. $2\left(\dfrac{5}{6}\right)^5$
C. Probability of at least 4 successes	III. $\left(\dfrac{1}{6}\right)^5$
D. Probability of at most one success	IV. $\left(\dfrac{5}{6}\right)^5$

Choose the correct answer from the options given below:

2022: 25 May Shift 1

Match List I with List II

LIST I	LIST II
A. The variance of a Poisson distribution with mean $\lambda$ is	I. $\sqrt{\lambda}$
B. The standard deviation of a Poisson distribution with mean $\lambda$ is	II. 4
C. In a Poisson distribution, if mean is 4, then the standard deviation is	III. $\lambda$
D. In a Poisson distribution, if mean is 4, then the variance is	IV. 2

Choose the correct answer from the options given below:

2026: 30 May Shift 1

On a multiple choice examination with four possible answers (out of which only one is correct) for each of five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

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

Solution

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