Question:

A dice is thrown twice. Then the probability of that '5' will come up at least once.

Updated On: Apr 17, 2025
  • \(\frac{11}{36}\)
  • \(\frac{25}{36}\)
  • \(\frac{23}{36}\)
  • \(\frac{12}{36}\)
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The Correct Option is A

Solution and Explanation

To solve the problem, we need to find the probability that a '5' will come up at least once when a fair six-sided die is thrown twice. We can approach this problem using the complementary probability method.

Step 1: Understand the total outcomes.
When a die is thrown twice, there are \(6 \times 6 = 36\) possible outcomes, since each throw has 6 possible results.

Step 2: Calculate the probability of not getting a '5' in a single throw.
The probability of not getting a '5' in a single throw is:

\[ P(\text{not } 5) = \frac{5}{6} \]

Step 3: Calculate the probability of not getting a '5' in both throws.
If the die is thrown twice and we do not get a '5' in either throw, the probability is the product of the probabilities of not getting a '5' in each throw (since the throws are independent):

\[ P(\text{not } 5 \text{ in both throws}) = P(\text{not } 5) \times P(\text{not } 5) = \left(\frac{5}{6}\right) \times \left(\frac{5}{6}\right) = \frac{25}{36} \]

Step 4: Use the complementary probability.
The probability of getting at least one '5' is the complement of the probability of not getting a '5' in both throws. Therefore:

\[ P(\text{at least one } 5) = 1 - P(\text{not } 5 \text{ in both throws}) = 1 - \frac{25}{36} = \frac{36}{36} - \frac{25}{36} = \frac{11}{36} \]

Final Answer:
The probability that a '5' will come up at least once is \(\frac{11}{36}\).

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