Question

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. In a binomial distribution, if n = 20, q = 0.75 then its mean is	I. 25
B. An unbiased coin is tossed 4 times. The mean of the number of tails is	II. 5
C. In a binomial distribution, mean is 5 and variance is 4, then the number of trials is	III. 0.5
D. In a binomial distribution, if mean is 8 and variance is 4, then the probability of success is	IV. 2

Choose the correct answer from the options given below:

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

Hint:

Mean is np and variance is npq. Divide variance by mean to get q.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In a binomial distribution with n trials and success probability p, the mean is np and the variance is npq, where q = 1 - p.

Step 2: Match A.
Given \(n=20\) and \(q=0.75\), so \(p=1-0.75=0.25\).
\[ \text{Mean} = np = 20 \times 0.25 = 5 \] So A matches II.

Step 3: Match B.
An unbiased coin has p = 0.5 for tails. For 4 tosses: \[ np = 4 \times 0.5 = 2 \] So B matches IV.

Step 4: Match C.
Mean np = 5 and variance npq = 4. Divide the variance by the mean: \[ q = \frac{4}{5} = 0.8 \] so \(p = 0.2\). Then \[ n = \frac{5}{0.2} = 25 \] So C matches I.

Step 5: Match D.
Mean np = 8 and variance npq = 4. Divide: \[ q = \frac{4}{8} = 0.5 \] so \(p = 1 - 0.5 = 0.5\). So D matches III.

Step 6: Choose the option.
The pairs are A-II, B-IV, C-I, D-III. This is option 3. Options 1, 2 and 4 each have at least one wrong pair, for example A-III in option 1 (mean 0.5 is not 5) and B-I in options 2 and 4 (25 is not the mean of 4 tosses).

Final Answer:
The matching is A-II, B-IV, C-I, D-III. \[ \boxed{\text{Option (3)}} \]