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Question
A die is tossed thrice. Find the probability of getting an odd number at least once.
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Solution
The total number obtained in the first throw of the dice = 6
And the case of not getting an odd number = 3
∴ Probability of not getting an odd number in the first toss P(A) = `3/6 = 1/2`
Similarly, the probability of not getting an odd number in the second toss is P(B) = `1/2`
Probability of not getting an odd number in the third toss P(C) = `1/2`
∵ The above three events are independent.
∴ The probability of all three happening together i.e. the event of not getting an odd number in each toss
P(A ∩ B ∩ C) = P(A) . P(B) . P(C)
= `1/2 . 1/2 . 1/2`
= `1/8`
∴ Probability of getting an odd number at least once
= 1 − P(A ∩ B ∩ C)
= `1 - 1/8`
= `7/8`
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