Question

advice is tossed once .
find the probability of getting a number greater than 4​

Answer 1

Answer:

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Step-by-step explanation:

Given :

A die is thrown once .

A die has 6 faces marked as 1, 2, 3, 4, 5 and 6.

If we throw one die then there possible outcomes are as follows: 1, 2, 3, 4, 5 and 6

Number of possible outcomes are = 6

Let E = Event of getting a number greater than 4

Number greater than 4 on a die are : 5,6

Number of outcome favourable to E = 2

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E) = 2/6 = 1/3

Hence, the required probability of getting a number greater than 4, P(E) = 1/3

Answer 2

Answer:

Given :

A die is thrown once .

A die has 6 faces marked as 1, 2, 3, 4, 5 and 6.

If we throw one die then there possible outcomes are as follows: 1, 2, 3, 4, 5 and 6

Number of possible outcomes are = 6

Let E = Event of getting a number greater than 4

Number greater than 4 on a die are : 5,6

Number of outcome favourable to E = 2

Probability (E) = Number of favourable outcomes / Total number of outcomes

P(E) = 2/6 = 1/3

Hence, the required probability of getting a number greater than 4, P(E) = 1/3