Question:
A die is thrown four times. The probability of getting a perfect square in at least one throw is
A die is thrown four times. The probability of getting a perfect square in at least one throw is
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Whenever a probability question contains the phrase "at least one," always calculate the probability of the opposite "none" condition first. Subtracting the chance of total failure from 1 bypasses summing up multiple separate successful cases and leads straight to the answer.
Updated On: Jun 11, 2026
- $\frac{58}{61}$
- $\frac{16}{81}$
- $\frac{65}{81}$
- $\frac{23}{81}$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
A standard six-sided fair die is rolled exactly four independent times. We need to compute the probability of rolling a perfect square number in at least one of those four throws.
Step 2: Key Formula or Approach:
The phrase "at least one" indicates that we can solve this problem efficiently using complementary probability: $$P(\text{At least one perfect square}) = 1 - P(\text{No perfect squares in any throw})$$ Since each die roll is independent, the probability of an event failing across all four attempts is simply the individual failure probability raised to the fourth power: $P(\text{None}) = [P(\text{Failure})]^4$.
Step 3: Detailed Explanation:
Let's analyze the sample space of a single die roll: $S = \{1, 2, 3, 4, 5, 6\}$.
Identify the numbers in this set that are perfect squares:
• $1 = 1^2$ (Perfect square)
• $4 = 2^2$ (Perfect square)
There are exactly 2 perfect squares out of 6 possible outcomes. Thus, the probability of getting a perfect square on a single throw ($p$) is: $$p = \frac{2}{6} = \frac{1}{3}$$ The probability of NOT getting a perfect square on a single throw ($q$) is: $$q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$$ Now find the probability of rolling zero perfect squares across all four independent attempts: $$P(\text{No perfect square in 4 throws}) = \left(\frac{2}{3}\right)^4 = \frac{16}{81}$$ Finally, subtract this complementary value from 1 to find our target probability: $$P(\text{At least one perfect square}) = 1 - \frac{16}{81} = \frac{81 - 16}{81} = \frac{65}{81}$$ This matches option (C).
Step 4: Final Answer:
The probability of getting a perfect square in at least one throw is $\frac{65}{81}$, which corresponds to option (C).
A standard six-sided fair die is rolled exactly four independent times. We need to compute the probability of rolling a perfect square number in at least one of those four throws.
Step 2: Key Formula or Approach:
The phrase "at least one" indicates that we can solve this problem efficiently using complementary probability: $$P(\text{At least one perfect square}) = 1 - P(\text{No perfect squares in any throw})$$ Since each die roll is independent, the probability of an event failing across all four attempts is simply the individual failure probability raised to the fourth power: $P(\text{None}) = [P(\text{Failure})]^4$.
Step 3: Detailed Explanation:
Let's analyze the sample space of a single die roll: $S = \{1, 2, 3, 4, 5, 6\}$.
Identify the numbers in this set that are perfect squares:
• $1 = 1^2$ (Perfect square)
• $4 = 2^2$ (Perfect square)
There are exactly 2 perfect squares out of 6 possible outcomes. Thus, the probability of getting a perfect square on a single throw ($p$) is: $$p = \frac{2}{6} = \frac{1}{3}$$ The probability of NOT getting a perfect square on a single throw ($q$) is: $$q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$$ Now find the probability of rolling zero perfect squares across all four independent attempts: $$P(\text{No perfect square in 4 throws}) = \left(\frac{2}{3}\right)^4 = \frac{16}{81}$$ Finally, subtract this complementary value from 1 to find our target probability: $$P(\text{At least one perfect square}) = 1 - \frac{16}{81} = \frac{81 - 16}{81} = \frac{65}{81}$$ This matches option (C).
Step 4: Final Answer:
The probability of getting a perfect square in at least one throw is $\frac{65}{81}$, which corresponds to option (C).
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