Question

Two dice are rolled simultaneously. The probability that they show different faces is
(a)$\frac{2}{3}$
(b)$\frac{1}{3}$
(c)$\frac{1}{3}$
(d)$\frac{5}{6}$

Answer 1

SOLUTION :

The correct option is (d) : 5/6

Given : Two dice are rolled simultaneously .

If we throw two dices then there possible outcomes are as follows:

{(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)}

Total no. of possible outcomes when 2 dice are thrown = 6 x 6 = 36

Total Number of possible outcomes = 36

Let E = Event of getting different faces on both dice

Favorable outcomes(different faces on both dice) :  

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

Number of favorable outcomes = 30

Probability ,P(E) = Number of favourable outcomes / total number of outcomes

P(E1) = 30/36 = 5/6

Hence, the Probability of getting different faces on both dice,  P(E1) = 5/6

HOPE THIS ANSWER WILL HELP  YOU…

Answer 2

Hi there!
Here's the answer:

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Given,
Two dice are rolled simultaneously

Let S be Sample space
n(S) - No. of total Outcomes when 2 dice are rolled

n(S) = 6² = 36

S =
{(1,1) , (1,2) ... (1,6),
(2,1) , (2,2) ... (2,6),
(3,1) , (3,2) ... (3,6),
(4,1) , (4,2) ... (4,6),
(5,1) , (5,2) ... (5,6),
(6,1) , (6,2) ... (6,6)}

Let E be the Event that both the dice show up different Number

Let E' be the Event that both the dice show up the same Number.

n(E') - No. of favorable outcomes for occurrence of E'

E' = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

n(E') is the no. of elements in set E'

•°• n(E') = 6

We have,

E + E' = S
=> n(E) + n(E') = n(S)
=> n(E) = n(S) - n(E')

=> No. of favorable outcomes for occurrence of E, n(E) = 36-6 = 30

$Probability = \dfrac{No.\: of\: favorable\: outcomes}{Total\: No.\: of\: outcomes}$

$P(E) = \dfrac{n(E)}{n(S)}$

$P(E) = \frac{30}{36} = \frac{5}{6}$

•°• Required Probability = $\frac{5}{6}$

This answer exists in Option (d)

•°• Option (d) is Correct.

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