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Question
A die is thrown, find the probability of following events:
- A prime number will appear,
- A number greater than or equal to 3 will appear,
- A number less than or equal to one will appear,
- A number more than 6 will appear,
- A number less than 6 will appear.
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Solution
Sample space of a experiment in throwing a dice
S = {1, 2, 3, 4, 5, 6}
That is, total possible outcomes n(S) = 6
i. Prime numbers are 2, 3, 5.
n (A) = 3
Hence, probability of a prime number appearing = `("n"("A"))/("n"("S")) = 3/6 = 1/2`
ii. Let the event 3 or a number greater than 3 be denoted by B, 3 or numbers greater than 3 are 3, 4, 5, 6.
n (B) = 4
Hence, probability, P(B) = `("n"("B"))/("n"("s")) = 4/6 = 2/3`
iii. Let the event 1 or a number less than 1 be denoted by C.
Numbers greater than 1 or 1 = 1
∴ n(C) = 1
Hence, probability, P(C) = `1/6`
iv. There is no number greater than 6 on a die, i.e. its probability = `0/6 = 0`
v. Numbers less than 6 are: 1, 2, 3, 4, 5. If it is represented by E, then
n(E) = 5
Hence, probability, P(E) = `5/6`
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