Question

when a die is thrown, find the probability of getting no.greater than 4
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Answer 1

Answer:

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

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

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Answer 2

Answer:

Let,E be the event getting no greater than 4

E={5,6}

n(E)=2

sample space(S)={1,2,3,4,5,6}

n(S)=6

probability of the given event(E)=p(E)=n(E)/n(S)=2/6=1/3