Question:
A coin whose faces are marked as \(5\) and \(3\) is tossed \(4\) times. The probability that the sum of the numbers on the coin is greater than \(17\) is ___
A coin whose faces are marked as \(5\) and \(3\) is tossed \(4\) times. The probability that the sum of the numbers on the coin is greater than \(17\) is ___
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Let k be the number of times 5 appears; express the sum in terms of k and find which values of k satisfy the condition.
Updated On: Nov 12, 2026
- \(\frac{5}{8}\)
- \(\frac{1}{16}\)
- \(\frac{5}{16}\)
- \(\frac{11}{16}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Let \(k\) be the number of tosses that show 5. The remaining \(4-k\) tosses show 3.
Step 2: Sum
Sum \(=5k+3(4-k)=2k+12\).
Step 3: Condition
\(2k+12>17\Rightarrow k>2.5\), so \(k=3\) or \(4\).
Step 4: Probability
Each outcome has probability \(\dfrac1{16}\) in total, with \(\binom43+\binom44=4+1=5\) favourable ways.
So \(P=\dfrac5{16}\).
Final Answer:
The probability is \(\dfrac5{16}\). \[ \boxed{\text{(C) }\dfrac{5}{16}} \]
Let \(k\) be the number of tosses that show 5. The remaining \(4-k\) tosses show 3.
Step 2: Sum
Sum \(=5k+3(4-k)=2k+12\).
Step 3: Condition
\(2k+12>17\Rightarrow k>2.5\), so \(k=3\) or \(4\).
Step 4: Probability
Each outcome has probability \(\dfrac1{16}\) in total, with \(\binom43+\binom44=4+1=5\) favourable ways.
So \(P=\dfrac5{16}\).
Final Answer:
The probability is \(\dfrac5{16}\). \[ \boxed{\text{(C) }\dfrac{5}{16}} \]
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