Question

A pair of dices thrown once fin probablity of getting same number on each dice

Answer 1

Number of possible outcomes are :-

(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)

= 36 

Total no of outcomes = 36

Getting same number when a dice is thrown for a certain event (A) = 6

That is (1,1) (2,2) (3,3) (4,4) (5,5) and (6,6)

Therefore p(A) = 6/36 = 1/6

Hope you like it please mark as brainliest

Answer 2

$\boxed{\boxed{\mathbf{\frac{1}{6}}}}$
_________________________________

$\underline{\underline{\Huge\mathfrak{ANSWER :}}}$


$\underline\text{Using the Formula,}$

$\boxed{\mathsf{P(E) = \frac{Number \: of \: Favourable \: Outcomes}{Total \: Number \: of \: Outcomes} }}$

====================================

POSSIBLE OUTCOMES ARE :-

(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)

= $\bf{36}$

====================================

FAVOURABLE OUTCOMES :-

(1,1) (2,2) (3,3)
(4,4) (5,5) (6,6)

= $\bf{6}$

====================================

Now,

$\underline\text{Putting the values in the Formula,} \\\underline\text{We get,}$

P(E) = $\large\mathsf{\frac{\cancel{6}}{\cancel{36}}}$

=> P(E) = $\large\mathsf{\frac{1}{6}}$

Hence, the probability to get even numbers on both dice is $\large\mathsf{\frac{1}{6}}$
_________________________________

$\boxed{\boxed{\boxed{\Large\mathsf{THANKS}}}}$

$\huge\mathbf{\# \: Be Brainly}$