MCQ
Two dice are thrown together. The probability that neither they show equal digits nor the sum of their digits is 9 will be:
- A$\frac{13}{15}$
- ✓$\frac{13}{18}$
- C$\frac{1}{9}$
- D$\frac{8}{9}$
✓
Answer
Correct option: B.
$\frac{13}{18}$
When two dices are thrown, there are (6 × 6) = 36 outcomes.
The set of all these outcomes is the sample space is given by
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)
Then E' = event of getting either a doublet or a total of 9
Thus, E' = {{1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (3, 6), (4, 5), (5, 4), (6, 3)}
i.e. n(E') = 10
$\text{P(E}')=\frac{\text{n(E}')}{\text{n(E)}}=\frac{10}{36}=\frac{5}{18}$
Hence, required probability P(E) = 1 - P(E')
$=1-\frac{5}{18}=\frac{13}{18}$
The set of all these outcomes is the sample space is given by
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)
$\therefore\text{n}\text{(S)} = 36$
Let E be the event of getting the digits which are neither equal nor give a total of 9.Then E' = event of getting either a doublet or a total of 9
Thus, E' = {{1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (3, 6), (4, 5), (5, 4), (6, 3)}
i.e. n(E') = 10
$\text{P(E}')=\frac{\text{n(E}')}{\text{n(E)}}=\frac{10}{36}=\frac{5}{18}$
Hence, required probability P(E) = 1 - P(E')
$=1-\frac{5}{18}=\frac{13}{18}$
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