Question

one number is selected at random from the first two hundred positive integers. what is the probability that the selected number is divisible by 6 or 8?




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Answer 1

Answer:

Answer: The probability of choosing a number that is divisible by 8 or 6 from the first 200 positive integers is 0.27.

Step-by-step explanation:

hope it helps you

Answer 2

Step-by-step explanation:

Total Number of cases = 200

Let A be the integer which is divisible by 6There are 33 integers from first 200 numbers, which are divisible by 6

$\boxed{{\tt \: hence\: , P \: (A) \: = \frac{33}{200} }}$

Let B be the integer which is divisible by 8There are 25 integers from first 200 numbers, which are divisible by 8

$\boxed{\tt \: Hence, P(A) = \frac{25}{200} }$

A∩B be the event such that an integer is divisible by both 6 and 8,

There are 8 integers from the first 200 numbers, which are divisible by both 6 an

$\boxed{\tt \: P(A∩B) = 8/200}$

Hence, the probability that an integer chosen is divisible by 6 or 8 isP(AUB) = P(A) + P(B)-P(A∩B)Now, substitute the values in the above form, we get

$\sf \: P(AUB) \: = ( \frac{33}{200} ) + ( \frac{25}{200} ) – ( \frac{8}{200} ) \\ \sf P(AUB) = \frac{50}{200} = \frac{1}{4}$

$\boxed{ \bf \red{ \therefore \: \frac{1}{4} is \: the \: required \: answer}}$