Question

find probability of getting a number divisible by 3 or a square number when a fair dice is thrown​

Answer 1

Answer:

When a fair die is thrown, the probability of getting each of the digit 1,2,3,4,5 and 6 is 16 . Of these, digits divisible by 3 are 3 and 6. Hence when you roll a fair die once, probability of getting a number divisible by 3 is 26 or 13 . Since the die is rolled 3 times, tje required probability is (23)3 or 827 .

Answer 2

Answer:

Concept:

Probability is a branch of mathematics involved with mathematical explanations of the chances of something happening or the truth of a statement. A number among 0 and 1, with 0 signifying impossibility and 1 representing certainty, represents the likelihood of an event. The higher the likelihood of an occurrence, the more probable it is to take place. The two possibilities ("heads" and "tails") are equally likely since the coin is fair; the probability of "heads" equals the probability of "tails"; and because no other outcomes are possible, the probability of either "heads" or "tails" is 1/2.

Given:

When a fair dice is thrown, find the probability of getting a number divisible by three or a square number.

Find:

find the answer for the given question

Answer:

Let $A$ represent the event of numbers divisible by three.

and the square number event is $B$.

Then,

{${1, 2, 3, 4, 5, 6}$} are possible outcomes ($S$)

numbers divisible by $3$  = {$3, 6$}

square numbers ($B$) = {$1, 4$}

∴ $n(S)=6, n(A)=2$  $and$  $n(B)=2$

To find: $n(A$∪$B)$

Here,

A and B are mutual events.

Then,

$P(A$∪$B)$  $= p(A) + p(B)$

              $=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}$

              $=\frac{2}{6}+\frac{2}{6}$

              $=\frac{2}{3}$

∴ The chances of getting a three-digit number or a square number is $\frac{2}{3}$