Question

Rolling two fair dice the probability of getting equal number on numbers with an even product is

Answer 1

$\huge\mathbb\fcolorbox{purple}{lavenderblush}{✰αηsωεя࿐}$

Each dice has a 1/2 probability of being odd hence P(odd)=1/2*1/2=1/4. So the probability of getting an even product is 1–1/4=3/4.

Answer 2

Answer:

The required answer is $\frac{1}{12}$.

Step-by-step explanation:

The sample space of rolling the two dice is

$\left\{\begin{array}{cccccc}(1,1)&(1,1)&(1,2)&(1,3)&(1,4)&(1,5)\\(2,1)&(2,2)&(2,3)&(2,4)&(2,5)&(2,6)\\(3,1)&(3,2)&(3,3)&(3,4)&(3,5)&(3,6)\\(4,1)&(4,2)&(4,3)&(4,4)&(4,5)&(4,6)\\(5,1)&(5,2)&(5,3)&(5,4)&(5,5)&(5,6)\\(6,1)&(6,2)&(6,3)&(6,4)&(6,5)&(6,6)\end{array}\right\}$

Total number of outcomes = $6^2$

                                              = 36

Consider the space of getting equal numbers.

$\{ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$

Now,

a) $(1,1)$

Product of the numbers = $1\times 1$

                                        = 1, an odd number

b) $(2,2)$

Product of the numbers = $2\times 2$

                                        = 4, an even number

c) $(3,3)$

Product of the numbers = $3\times 3$

                                        = 9, an odd number

d) $(4,4)$

Product of the numbers = $4\times 4$

                                        = 16, an even number

e) $(5,5)$

Product of the numbers = $5\times 5$

                                        = 25, an odd number

f) $(6,6)$

Product of the numbers = $6\times 6$

                                        = 36, an even number

Total no. of getting equal numbers with an even product = 3

i.e., number of favorable outcomes = 3

Probability of getting equal numbers with an even product,

= $\frac{Number\ of\ favorable\ outcomes}{Total\ number\ of\ outcomes}$

= $\frac{3}{36}$

= $\frac{1}{12}$

#SPJ3