Find the probability that a number from 1 to 300 is divisible by 3 or 7 ?
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Total number of integer numbers given: 300
Let X be a number randomly chosen from 1 to 300.
Numbers divisible by 3: 3 * 1, 3 * 2, ..... 3 * 100 : so there are 100 of them.
Numbers divisible by 7: 7 * 1, 7*2, .... 7*42 = 294
so there are 42 of them.
Numbers divisible by both 3 and 7: ie, by 21: 21*1, 21*2, ... 21*14=294
so there are 14 of them.
Now we have totally 100 + 42 - 14 = 128 integers from 1 to 300 which are divisible by 3 or 7.
Probability = 128 / 300 = 32/75
Let X be a number randomly chosen from 1 to 300.
Numbers divisible by 3: 3 * 1, 3 * 2, ..... 3 * 100 : so there are 100 of them.
Numbers divisible by 7: 7 * 1, 7*2, .... 7*42 = 294
so there are 42 of them.
Numbers divisible by both 3 and 7: ie, by 21: 21*1, 21*2, ... 21*14=294
so there are 14 of them.
Now we have totally 100 + 42 - 14 = 128 integers from 1 to 300 which are divisible by 3 or 7.
Probability = 128 / 300 = 32/75
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total numbers divisible my 3 is 3,6,9,...,300
n = 300/3 = 100
total numbers divisible by 7
7,14,21,..., 294
294 = 7 + (n1-1)7
n1 = 42
numbers divisible by both 3 and 7
21,42,63,...,294
n2= 294/21 = 14
by principle of exclusion theory
total numbers = 100 + 42 - 14 = 128
probability = 128/300 = 32/75
n = 300/3 = 100
total numbers divisible by 7
7,14,21,..., 294
294 = 7 + (n1-1)7
n1 = 42
numbers divisible by both 3 and 7
21,42,63,...,294
n2= 294/21 = 14
by principle of exclusion theory
total numbers = 100 + 42 - 14 = 128
probability = 128/300 = 32/75
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