Suppose we throw a die once. (i) What is the probability of getting a

1.	Suppose we throw a die once. (i) What is the probability of getting a number greater than 4 ? (ii) What is the probability of getting a number less than or equal to 4 ?
Answer» (i) Here, let E be the event ‘getting a number greater than 4’. The number of possible outcomes is six : 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5 and 6. Therefore, the number of outcomes favourable to E is 2. So, P(E) = P(number greater than 4) =2/6 =1/3 (ii) Let F be the event ‘getting a number less than or equal to 4’. Number of possible outcomes = 6 Outcomes favourable to the event F are 1, 2, 3, 4. So, the number of outcomes favourable to F is 4. Therefore, P(F) =4/6 =2/3 Are the events E and F in the example above elementary events? No, they are not because the event E has 2 outcomes and the event F has 4 outcomes.

Suppose we throw a die once. (i) What is the probability of getting a number greater than 4 ? (ii) What is the probability of getting a number less than or equal to 4 ?

Answer»

(i) Here, let E be the event ‘getting a number greater than 4’. The number of possible outcomes is six : 1, 2, 3, 4, 5 and 6, and the outcomes favourable to E are 5 and 6. Therefore, the number of outcomes favourable to E is 2. So,

P(E) = P(number greater than 4) =2/6 =1/3

(ii) Let F be the event ‘getting a number less than or equal to 4’.

Number of possible outcomes = 6

Outcomes favourable to the event F are 1, 2, 3, 4.

So, the number of outcomes favourable to F is 4.

Therefore, P(F) =4/6 =2/3

Are the events E and F in the example above elementary events? No, they are not because the event E has 2 outcomes and the event F has 4 outcomes.

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