Question

An unbiased coin is tossed \(3\) times. If the third toss gets head, then the probability of getting at least one more head is

• A. \(\dfrac{3}{4}\)
• B. \(\dfrac{1}{4}\)
• C. \(\dfrac{1}{2}\)
• D. \(\dfrac{1}{3}\)

Hint:

In conditional probability questions, first restrict the sample space according to the given condition, then count favorable outcomes within that restricted sample space.

Answer 1

The Correct Option is A

Step 1: Fix the given condition.
A coin is tossed 3 times. We are told the third toss is Head (H). So the outcome of the third toss is fixed as H; only the first two tosses are random.
Step 2: List all outcomes consistent with the condition.
The first two tosses can each be H or T independently, giving 4 possibilities: HH, HT, TH, TT. Combined with the fixed third toss H, the four equally likely outcomes are: HHH, HTH, THH, TTH.
Step 3: Identify the event of interest.
We want 'at least one more head', meaning at least one head among the first two tosses (the third is already a head, so that head does not count as 'more').
Step 4: List favourable outcomes.
Among HH, HT, TH, TT (first two tosses), the ones with at least one H are HH, HT, TH. The corresponding full outcomes are HHH, HTH, THH. That is 3 favourable outcomes.
Step 5: Find the probability.
\[ P = \frac{\text{favourable}}{\text{total}} = \frac{3}{4} \]
Step 6: Cross-check using complementary counting.
The only unfavourable outcome (no head in first two tosses) is TTH -- just 1 case out of 4. So $P = 1 - \frac{1}{4} = \frac{3}{4}$. Both approaches agree.
Step 7: State the final answer.
\[ \boxed{\dfrac{3}{4}} \]