Question

Out of the number 1 to 100,one is selected at random.What is the probability that is divisible by 7 or 8

Answer 1

Answer:

6/25

Step-by-step explanation:

Out of the number 1 to 100,one is selected at random.What is the probability that is divisible by 7 or 8

1 to 100  - total number = 100

Number multiplied by 7  = 13

= ( 7 , 14 , 21 ............................, 91 , 98)

Number multiplied by 8  = 12

= ( 8 , 16 , 24 ............................, 88 , 96)

56 is common in both

total numbers divisible by 7 or 8 = 13 + 12 - 1 = 24

probability that is divisible by 7 or 8 = 24/100 = 6/25

Answer 2

Answer:

The required probability is

$\frac{1}{4}$

Step-by-step explanation:

Formula used:

Addition theorem on probability:

Let A and B be two events.

Then

P(AUB) = P(A)+P(B) - P(A∩B)

Here

S={1,2,.............100}

Let A be the event of getting number which is divisible by 7

A={7,14,21...........98}

Then

$P(A)=\frac{n(A)}{n(S)}$

$P(A)=\frac{14}{100}$

Let B be the event of getting number which is divisible by 8

B={8,16,24,........96}

Then

$P(B)=\frac{n(B)}{n(S)}$

$P(B)=\frac{12}{100}$

A∩B={56}

P(A∩B)=n(A∩B)/n(S)

P(A∩B)$=\frac{1}{100}$

Now

P(AUB) = P(A)+P(B) - P(A∩B)

P(AUB)$=\frac{14}{100}+\frac{12}{100}-\frac{1}{100}$

P(AUB)$=\frac{14+12-1}{100}$

P(AUB)$=\frac{25}{100}$

P(AUB)$=\frac{1}{4}$