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Mathematics

Assertion (A): A die is thrown twice. The probability of getting a bigger value on the first throw is (512)\Big(\dfrac{5}{12}\Big).

Reason (R): When we throw a die once, total number of possible outcomes is 6 and when the die is thrown twice, the total number of possible outcomes is 6 + 6 = 12.

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false

Probability

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Answer

When a dice is thrown twice,

Total number of outcomes = 36

Let E be the event of getting a bigger value on the first throw,

E = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}

The number of favorable outcomes of the event E = 15

P(E)=Number of favorable outcomesTotal number of outcomes=1536=512\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{15}{36} = \dfrac{5}{12}

Therefore, Assertion (A) is true

When a single die is thrown, there are 6 possible outcomes {1, 2, 3, 4, 5, 6}.

When a die is thrown twice, the total number of outcomes is determined by the product of the outcomes of each throw:

Total Outcomes = 6 × 6 = 36

Reason (R) is false.

A is true, R is false

Hence, option 1 is the correct option.

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