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(1/6)^5, B. Probability of 5 successes, II. 2(5/6)^5, C. Probability of at least 4 successes, III. (1/6)^5, D. Probability of at most one success, IV. (5/6)^5, Choose the correct answer from the options given below: Option 1: A-IV, B-II, C-I, D-III Option 2: A-IV, B-I, C-II, D-III Option 3: A-IV, B-III, C-II, D-I Option 4: A-IV, B-III, C-I, D-II

Answer: A-IV, B-III, C-I, D-II. Solution: A sum of 7 occurs in 6 of 36 ways, so p=1/6, q=5/6, n=5. No success: (5/6)^5 (IV). Five successes: (1/6)^5 (III). At least 4: 55/6^5+1/6^5 = 26(1/6)^5 (I). At most one: (5/6)^5+51/6(5/6)^4 = 2(5/6)^5 (II).

Applied Mathematics

Probability Distribution

Hard

applied

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:

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. $\left(\dfrac{5}{6}\right)^5$

A-IV, B-II, C-I, D-III

A-IV, B-I, C-II, D-III

A-IV, B-III, C-II, D-I

A-IV, B-III, C-I, D-II

✅ Correct Option: 4

A sum of 7 occurs in 6 of 36 ways, so $p=\dfrac{1}{6}$ , $q=\dfrac{5}{6}$ , $n=5$ .

No success: $\left(\dfrac{5}{6}\right)^5$ (IV). Five successes: $\left(\dfrac{1}{6}\right)^5$ (III).

At least 4: $5\cdot\dfrac{5}{6^5}+\dfrac{1}{6^5} = 26\left(\dfrac{1}{6}\right)^5$ (I).

At most one: $\left(\dfrac{5}{6}\right)^5+5\cdot\dfrac{1}{6}\left(\dfrac{5}{6}\right)^4 = 2\left(\dfrac{5}{6}\right)^5$ (II).

2023: 23 May Shift 3

A book consisting of 2000 pages has 540 misprints distributed randomly throughout the book. The average number of misprints in one page of the book is:

2026: 25 May Shift 1

If $X$ is a random variable such that $X$ can take values 0,1, 2 or 3. Then the expectation of $X$ for the following data is :

$X$	0	1	2	3
$P(X)$	$k$	$k^2$	$1-5k^2$	$k^2$

(where $k>0$ )

2026: 22 May Shift 2

The probability that a man aged 35 years will die before reaching the age of 40 years may be taken as 0.018. Out of a group of 100 men, now aged 35 years, what is the approximate probability that one man will die within next five years?

(Given: $e^{-1.8}=0.1653$ )

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: 7 June Shift 1

If the mean of a binomial distribution is 24 and its standard deviation is 4 , then the probability of getting success is :

2023: 15 June Shift 2

A telephone exchange receives on an average 5 calls per minute. The probability of receiving 3 or less calls per minute is :

Solution

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