Math, asked by faris631786, 3 months ago

A die is thrown once, find the probability of getting
lying
between 2 and 6.
(iii) an odd number
(1) a number greater than 6.
composite number
(i) a prime number
(i) a number greater than 6
(v) a composite number ​

Answers

Answered by TheLightofWisdom
0

Answer:

(i) odd numbers between 2 and 6 = 3, 5

total numbers = 6

required probability = no. of favourable outcomes/total no. of outcomes = 2/6 = 1/3

(ii) Can there be a no. greater than 6 on the die? No! so favourable outcomes = 0, and thus probability = 0

(iii) Composite numbers between 2 and 6 = 4

(Composite numbers are those which have atleast a divisor other than 1 and itself, for example, 3 is NOT composite, as it's only divisors are 3 and 1. But 4 is a composite no. as the divisors of 4 are = 1,2,4. Also, 6 is a composite number, but we can not include it, as we only have to select numbers between 2 and 6.)

so, probability = 1/6

(iv) Prime numbers between 2 and 6 = 3,5

(Prime numbers are those numbers which are greater than 1 and have only 1 and itself as a divisor. Basically, all non-composite numbers are prime. But 1 is a unique number, not prime.)

so, probability = 2/6 = 1/3

There are all your answers! I hope it helps you understand the questions you've asked.

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