Question

Given below are two statements:
Statement I: A coin is tossed three times. The probability of getting exactly two heads is \(\frac{3}{8}\).
Statement II: In tossing of 10 coins, the probability of getting exactly 5 heads is \(\frac{63}{256}\).
In the light of the above statements, choose the correct answer from the options given below.

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Answer 1

The Correct Option is A

To determine the correctness of the given statements, we need to calculate the probabilities as mentioned in each statement and compare them with the provided values.

Statement I: A coin is tossed three times. The probability of getting exactly two heads.

The number of total outcomes when a coin is tossed three times is \(2^3 = 8\).

We need to find the number of ways to get exactly two heads. This can occur in the combinations: HHT, HTH, and THH.

The number of favorable outcomes is 3.

Thus, the probability is given by:

P(\text{exactly two heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{3}{8}

The statement is correct as the calculated probability matches the given probability \(\frac{3}{8}\).

Statement II: In tossing of 10 coins, the probability of getting exactly 5 heads is \(\frac{63}{256}\).

We use the binomial probability formula for this calculation:

P(\text{exactly } k \text{ heads}) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}

Where \(n=10\), \(k=5\), and \(p=\frac{1}{2}\).

Calculate \( \binom{10}{5} \):

\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252

Now, calculate the probability:

P(\text{exactly 5 heads}) = \binom{10}{5} \times \left( \frac{1}{2} \right)^5 \times \left( \frac{1}{2} \right)^5 = 252 \times \frac{1}{32} \times \frac{1}{32} = \frac{252}{1024} = \frac{63}{256}

The statement is correct as the calculated probability matches the given probability \(\frac{63}{256}\).

Thus, Both Statement I and Statement II are true.