Question

To win a game, Ruby wanted a total of 7.

What is the probability of winning a game by Ruby?

(i)

1
/6

(ii)

7
/12

(iii)

5
/18

(iv) 1
/9​

Answer 1

Answer:7 /12

Step-by-step explanation: It is 7 /12 as Ruby has to win 7 times to win the game so the numerator should be 7. Here only 7 /12 is the fraction that has 7 as a numerator, so the answer is 7 /12.

Answer 2

Complete question:

Ruby and Rita are best friends. They are staying in the same colony. Both are studying in the same class and in the same school. During Winter vacation Ruby visited Rita's house to play Ludo. They decided to play Ludo with 2 dice.

To win a game, Ruby wanted a total of 7.

What is the probability of winning a game by Ruby?

(i) 1/6

(ii) 7/12

(iii) 5/18

(iv) 1/9

Answer:

The probability of winning a game by Ruby is 1/6

Step-by-step explanation:

Given,

Ruby and Rita play Ludo using two dice.

If the sum of the numbers shown on the dice is 7, then Ruby wins the game.

To find,

The probability of winning a game by Ruby.

Calculation,

Let S be the sample space of throwing two dice. Then,

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)}

Hence, n(S) = 36

Now let E be the event of the sum shown on the dice is 7.

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

Hence, n(E) = 6

So, the probability of happening the event E is:

P(E) = n(E)/n(S)

⇒ P(E) = 6/36

⇒ P(E) 1/6

Therefore, the probability of winning a game by Ruby is 1/6.

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