Question

Two dice are thrown simultaneously.

The probability that sum is odd or less than $7$ or both is

• A. $\frac{2}{3}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C

$\frac{3}{4}$

Explanation for the correct option:

From the classical definition of probability we know,

Probability of an event $=\frac{\mathrm{number}\mathrm{of}\mathrm{favorable}\mathrm{outcomes}}{\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}$

If two dice are thrown simultaneously, the total number of sample space is $36$, which is as follows:

$$\begin{gathered} S=(1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) \end{gathered}$$

n(S)$=36$

Favorable outcomes(sum is odd)$=(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(3,6),(4,1),(4,3),(4,5),(5,2),(5,4),(5,6),(6,1),(6,3),(6,5)$

n(odd)$=18$

Favorable outcomes(sum less than 7)$=(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1)$

n(sum less than $7$)$=15$

Favorable outcomes(both)$=(1,2),(1,4),(2,1),(2,3),(3,2),(4,1)$

n(both)$=6$

The required probability $=$ P(sum is odd) $+$ P(sum less than $7$) $-$ P(both)

P(odd) $=\frac{18}{36}$

P(less than $7$)$=\frac{15}{36}$

P(both)$=\frac{6}{36}$

Required Probability $=\left(\frac{18}{36}\right)+\left(\frac{15}{36}\right)-\left(\frac{6}{36}\right)$

$$\begin{gathered} =\left(\frac{18+15-6}{36}\right) \\ =\frac{27}{36}=\frac{9}{12}=\frac{3}{4} \end{gathered}$$

Hence the correct option is option(C) i.e. $\frac{3}{4}.$