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
An unbiased coin is tossed \(3\) times. If the third toss gets head, then the probability of getting at least one more head is
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In conditional probability questions, first restrict the sample space according to the given condition, then count favorable outcomes within that restricted sample space.
Updated On: Jun 26, 2026
- \(\dfrac{3}{4}\)
- \(\dfrac{1}{4}\)
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{3}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understand the given condition.
A coin is tossed \(3\) times.
It is given that the third toss is head.
So the third toss is fixed as \[ H \]
Step 2: List possible outcomes for the first two tosses.
Since the third toss is already \(H\), the first two tosses can be: \[ HH,\ HT,\ TH,\ TT \] Thus, the possible outcomes are: \[ HHH,\ HTH,\ THH,\ TTH \]
Step 3: Count the total conditional outcomes.
There are \[ 4 \] possible outcomes when the third toss is fixed as head.
Step 4: Identify favorable outcomes.
We need at least one more head apart from the third toss head.
So among the first two tosses, at least one should be head.
Favorable outcomes are: \[ HHH,\ HTH,\ THH \]
Step 5: Count favorable outcomes.
The number of favorable outcomes is \[ 3 \]
Step 6: Find the probability.
\[ P=\frac{\text{Favorable outcomes}}{\text{Total outcomes}} \] \[ P=\frac{3}{4} \]
Step 7: Final conclusion.
Therefore, \[ \boxed{\frac{3}{4}} \]
A coin is tossed \(3\) times.
It is given that the third toss is head.
So the third toss is fixed as \[ H \]
Step 2: List possible outcomes for the first two tosses.
Since the third toss is already \(H\), the first two tosses can be: \[ HH,\ HT,\ TH,\ TT \] Thus, the possible outcomes are: \[ HHH,\ HTH,\ THH,\ TTH \]
Step 3: Count the total conditional outcomes.
There are \[ 4 \] possible outcomes when the third toss is fixed as head.
Step 4: Identify favorable outcomes.
We need at least one more head apart from the third toss head.
So among the first two tosses, at least one should be head.
Favorable outcomes are: \[ HHH,\ HTH,\ THH \]
Step 5: Count favorable outcomes.
The number of favorable outcomes is \[ 3 \]
Step 6: Find the probability.
\[ P=\frac{\text{Favorable outcomes}}{\text{Total outcomes}} \] \[ P=\frac{3}{4} \]
Step 7: Final conclusion.
Therefore, \[ \boxed{\frac{3}{4}} \]
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