All the students of a class were asked to write a three-digit number. A student was chosen at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.

The numbers given by the students was as follows: 837, 172, 643, 371, 124, 512, 432, 948, 311, 252, 999, 557, 784, 928, 867, 798, 665, 245, 107, 463, 267, 523, 314, 843, 350, 923, 264, 872, 712, 535, 726, 129, 347, 687, 621, 896, 512, 413, 247, 800.

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\frac{7}{20}

\frac{9}{20}

\frac{1}{20}

\frac{3}{7}

answer is A.

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Detailed Solution

Given data is 837, 172, 643, 371, 124, 512, 432, 948, 311, 252, 999, 557, 784, 928, 867, 798, 665, 245, 107, 463, 267, 523, 314, 843, 350, 923, 264, 872, 712, 535, 726, 129, 347, 687, 621, 896, 512, 413, 247, 800.
Probability of any event, P(E)

= \frac{n (E)}{n (S)}

Number of favourable outcomes = n(E); Number of total outcomes = n(S); Probability of event = P(E)
It is given that the total number of students is 40 so the total number of outcomes = n(S)= 40
The favourable outcome is getting a number divisible by 3, i.e., {837, 432, 948, 252, 999, 867, 798, 267, 843, 264, 726, 129, 687, 621}
Getting any of these will be a favourable outcome so, n(E)= 14
Probability of event occurring will be,