Question

Find the probability of getting an odd number on the first die or a total of 8 in a single throw of two dice

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Answer 1

Answer:

cases of getting f

odd number on first die=18

cases of getting a total of 8=5

total favourable cases=18+5=23

total cases=36

probability=23/36

Answer 2

Answer:

The probability of getting an odd number on the first die or a total of 8 in a single throw of two dice is $\frac{5}{9}$

Step-by-step explanation:

We know,

If A and B are mutually exclusive events associated with a random experiment having sample space S,then

$P(A \cup \: B) = P(A) + P(B)$

Here Number dice = 2

Outcomes of A,

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

$P(A) = \frac{18}{36}$

Favorable outcomes of B,

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

P(B) $= \frac{5}{36}$

Now,$n(A \cap \: B) = 3$

Required probability,

$P(A \cap \: B) \\ = \frac{n(P \cap \: B)}{n(s)} \\ = \frac{3}{36} \\ = \frac{1}{12}$

Again

$P(A \cup \: B) \\ = P(A) + P(B) - P(A \cap \: B) \\ = \frac{1}{2} + \frac{5}{36} - \frac{1}{12} \\ = \frac{18 + 5 - 3}{36} \\ = \frac{5}{9}$

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