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The probability of any event will be easier by using the standard formula which is mentioned in the question. An unbiased die is thrown, what is the probability of getting
(i).an even number or a multiple of 3
(ii). a number 3 or 4
(iii). a number greater than 3

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\frac{2}{3}

\frac{1}{3}

\frac{1}{2}

\frac{1}{2}

\frac{2}{6}

0

\frac{2}{3}

\frac{1}{6}

\frac{5}{6}

None

answer is A.

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Detailed Solution

Given, an unbiased die is thrown.
In this case the sample space,

S = \{1, 2, 3, 4, 5, 6\}

So overall or total outcomes =6
Now we know that, probability of an event,

P = \frac{No . of favorable outcomes}{Total no . of possible outcomes}

(i) In first part,
Total even number or multiple of 3 = 4 {2, 3, 4, 6}
Thus, favorable outcomes in this case= 4
So probability of even number or multiples of

3,

= \frac{4}{6}

= \frac{2}{3}

(ii) In second part we have a number 3 or 4
Total favorable outcomes=2
Thus, probability of a number 3 or 4,

= \frac{2}{6}

= \frac{1}{3}

(iii) In third part we have number greater than 3 {4, 5, 6}
So, total favorable outcomes=3
Now, probability of a no. greater than 3,

= \frac{3}{6}

= \frac{1}{2}

Option 1 is correct.

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