Question

Topic: probability
Question: Two dice are thrown and the points on them are added together. Find which is more likely to happen the sum is 7 and that the sum is 8.
Note: The nonsense answer is no appreciated and please leave it if you don't know the answer.
Reward: ill recharge your phone for rs.200 if your answer is appropriate.
Ans: Getting sum 7 is more likely to happen.
Good luck and thanking you.​

Answer 1

Step-by-step explanation:

The possible outcomes when a dice is thrown = {1, 2, 3, 4, 5, 6}

Number of possible outcomes of a dice = 6

(i) Prime numbers on a dice are 2, 3, and 5.

Total prime numbers on a dice = 3

Probability of getting a prime number =

(ii) Numbers lying between 2 and 6 = 3, 4, 5

Total numbers lying between 2 and 6 = 3

Probability of getting a number lying between 2 and 6

(iii) Odd numbers on a dice = 1, 3, and 5

Total odd numbers on a dice = 3

Probability of getting an odd number

Answer 2

$\large\underline{\sf{Solution-}}$

↝ When two dices are thrown, possible outcomes are

$\begin{gathered}\begin{gathered}\bf\: Sample \: space-\begin{cases} &\sf{11,12,13,14,15,16,} \\ &\sf{21,22,23,24,25,26,}\\ &\sf{31,32,33,34,35,36,}\\ &\sf{41,42,43,44,45,46,}\\ &\sf{51,52,53,54,55,56,}\\ &\sf{61,62,63,64,65,66} \end{cases}\end{gathered}\end{gathered}$

So, number of elements in sample space is

$\rm :\longmapsto\:n(S) = 36$

Case :- 1

Let A be the event getting sum as 7 on two dices.

So, favourable outcomes are

$\rm :\longmapsto\:A = \{16,61,25,52,43,34 \}$

$\rm :\implies\:n(A) = 6$

We know,

Probability of an event is given by

$\sf \:Probability \: of \: an \: event =\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes \: in \: sample \: space}$

$\rm :\longmapsto\:P(A) = \dfrac{n(A)}{n(S)}$

$\rm :\longmapsto\:P(A) = \dfrac{6}{36}$

$\bf :\longmapsto\:P(A) = \dfrac{1}{6}$

Case :- 2

Let B be the event getting sum as 8 when pair of dice is thrown.

So, possible favourable outcomes are

$\rm :\longmapsto\:B = \{26,62,44,35,53 \}$

$\rm :\implies\:n(B) = 5$

Now,

We know that

$\sf \:Probability \: of \: an \: event =\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes \: in \: sample \: space}$

$\rm :\longmapsto\:P(B) = \dfrac{n(B)}{n(S)}$

$\bf :\longmapsto\:P(B) = \dfrac{5}{36}$

From case 1 and case 2, we concluded that

$\rm :\longmapsto\:P(A) > P(B)$

$\bf\implies \:A \: is \: more \: likely \: to \: happen.$

Additional Information :-

The sample space is the collection of all possible outcomes associated with the random experiment.

An event associated with a random experiment is a part of the sample space.

The probability of an event, P (E) lies [0, 1].

The probability of sure event is 1.

The probability of impossible event is 0.

P(A) + P( not A) = 1