A fair coin is tossed 10 times. What is the probability that ONLY the first two tosses will yield heads?a)(1/2)2b)10C2 (1/2)3c)(1/2)10d)10C2 (1/2)10Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Let A be the event that first toss is head
And B be the event that second toss is head.

By the given condition rest all 8 tosses should be tail
∴ The probability of getting head in first two cases
=

Answer

Let A be the event that first toss is head
And B be the event that second toss is head.

By the given condition rest all 8 tosses should be tail
∴ The probability of getting head in first two cases
=

Answer

To find the probability that only the first two tosses will yield heads, we need to consider the possible outcomes of the remaining 8 tosses.

Total number of outcomes when tossing a fair coin 10 times:
Since each toss has 2 possible outcomes (heads or tails), the total number of outcomes when tossing a fair coin 10 times is 2^10 = 1024.

Number of outcomes with only the first two tosses yielding heads:
For only the first two tosses to yield heads, we need to fix the outcomes of the first two tosses as heads and allow the remaining 8 tosses to have any outcome (heads or tails).

The number of outcomes for the remaining 8 tosses is 2^8 = 256, as each toss has 2 possible outcomes and there are 8 tosses.

Therefore, the number of outcomes with only the first two tosses yielding heads is 1 * 1 * 2^8 = 2^8 = 256.

Probability of only the first two tosses yielding heads:
To find the probability, we divide the number of favorable outcomes (only the first two tosses yielding heads) by the total number of possible outcomes.

Probability = Number of favorable outcomes / Total number of possible outcomes

Probability = 256 / 1024

Simplifying, we get:

Probability = 1 / 4

Since 1/4 can be written as (1/2)^2, the correct answer is option 'C': (1/2)^10.

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