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A die is thrown, find the probability of following events: A prime number will appear, A number greater than or equal to 3 will appear, A number less than or equal to one will appear,

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Question

A die is thrown, find the probability of following events:

  1. A prime number will appear,
  2. A number greater than or equal to 3 will appear,
  3. A number less than or equal to one will appear,
  4. A number more than 6 will appear,
  5. A number less than 6 will appear.
Sum
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Solution

Sample space of a experiment in throwing a dice

S = {1, 2, 3, 4, 5, 6}

That is, total possible outcomes n(S) = 6

i. Prime numbers are 2, 3, 5.

n (A) = 3

Hence, probability of a prime number appearing = `("n"("A"))/("n"("S")) = 3/6 = 1/2`

ii. Let the event 3 or a number greater than 3 be denoted by B, 3 or numbers greater than 3 are 3, 4, 5, 6.

n (B) = 4

Hence, probability, P(B) = `("n"("B"))/("n"("s")) = 4/6 = 2/3`

iii. Let the event 1 or a number less than 1 be denoted by C.

Numbers greater than 1 or 1 = 1

∴ n(C) = 1

Hence, probability, P(C) = `1/6`

iv. There is no number greater than 6 on a die, i.e. its probability = `0/6 = 0`

v. Numbers less than 6 are: 1, 2, 3, 4, 5. If it is represented by E, then

n(E) = 5

Hence, probability, P(E) = `5/6`

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Chapter 14: Probability - EXERCISE 14.2 [Page 306]

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NCERT Mathematics [English] Class 11
Chapter 14 Probability
EXERCISE 14.2 | Q 3. | Page 306

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