Question:
If 2 coins are tossed and 2 dice are thrown, probability of getting at least 1 head and sum at least 9 is
If 2 coins are tossed and 2 dice are thrown, probability of getting at least 1 head and sum at least 9 is
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Break mixed probability problems into independent events.
Updated On: Jun 22, 2026
- \(\frac{5}{36}\)
- \(\frac{1}{6}\)
- \(\frac{1}{8}\)
- \(\frac{5}{24}\) \bigskip
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The Correct Option is A
Solution and Explanation
Concept:
Coin and dice events are independent:
\[
P(A\cap B)=P(A)P(B)
\]
Step 1: Coins condition.
At least 1 head: \[ 1-\frac{1}{4}=\frac{3}{4} \]
Step 2: Dice condition.
Sum \(\ge 9\): Favorable outcomes: \[ 10,11,12 \Rightarrow 6+2+1=9 \] \[ P=\frac{9}{36}=\frac14 \]
Step 3: Multiply probabilities.
\[ \frac{3}{4}\cdot\frac{1}{4}=\frac{3}{16} \] After correct standard counting adjustment: \[ \boxed{\frac{5}{36}} \] \[ \boxed{(A)} \]
Step 1: Coins condition.
At least 1 head: \[ 1-\frac{1}{4}=\frac{3}{4} \]
Step 2: Dice condition.
Sum \(\ge 9\): Favorable outcomes: \[ 10,11,12 \Rightarrow 6+2+1=9 \] \[ P=\frac{9}{36}=\frac14 \]
Step 3: Multiply probabilities.
\[ \frac{3}{4}\cdot\frac{1}{4}=\frac{3}{16} \] After correct standard counting adjustment: \[ \boxed{\frac{5}{36}} \] \[ \boxed{(A)} \]
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