Question

An integer is chosen from among the numbers 23 to 80. What is the probability that the number is divisible by 2 and 3 both?

Answer 1

Answer:

may it help you............

Answer 2

Probability will be equal to 0.666

Step-by-step explanation:

We have given number 23 to 80

Total count of number from 23 to 80 will be

24,25,26,27,28,29,30 = 7

31-40 = 10

41-50 = 10

51-60 =10

61-70 = 10

71-80 = 10

So total count of numbers from 23 to 80 = 10+10+10+10+10+7 = 57'

Number which are divisible by 2 are 24,26,28....80

Difference d = 2 first term a = 24

Last term = 80

Nth term is given by

$a+(n-1)d$

So $24+(n-1)\times 2=80$

$n=29$

Number which are divisible by 3 are 24,27,30,33,....78

So $78=24+(n-1)\times 3$

n = 19

The number which are divisible by both 2 and 3

24,30,36,42,48,54,60,66,72,78 = 10

So total number which are divisible by 2 and 3 are = 29+19-10 = 38

So probability will be equal to $=\frac{38}{57}=0.666$

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