Probability - SS2 Mathematics Past Questions and Answers - page 1
In the toss of a coin thrice, find the probability of three tails appearing.
\(probability\ of\ 3\ tails,\ P(TTT) = P(tail\ in\ 1st\ toss) + P(tail\ in\ 2nd\ toss) + P(tail\ in\ 3rd\ toss)\) (independent events)
\(P(TTT) = P(A \cap B \cap C) = P(A)P(B)P(C) = (\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) = \frac{1}{8}\)
\(A = \{ a,e,i,o,u\}\) and \(B = \{ p,h,f,v,m,w\}\). Find the probability of choosing a letter from the English alphabet which is either in event \(A\) or \(B\).
\[total\ number\ of\ english\ alphabets = 26\]
\[P(A) = \frac{5}{26}\]
\[P(B) = \frac{5}{26}\]
\(P(A\ or\ B) = P(A \cup B) = P(A) + P(B) = \ \frac{5}{26} + \frac{5}{26} = \frac{10}{26} = \frac{5}{13}\)
A ball is drawn at random from a box containing \(6\ red\ balls,\ 4\ white\ balls\ \)and\(\ 5\ green\ balls\). Determine the probability that it is:
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\(Red\)
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\(White\)
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\(Blue\)
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\(Not\ red\)
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\(Red\ or\ white\)
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\[total\ number\ of\ balls = 6 + 4 + 5 = 15\]
\[P(r) = \ \frac{6}{15},\ \ P(b) = \ \frac{4}{15},\ \ P(g) = \ \frac{5}{15}\]
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\(P(RED) = \frac{6}{15}\)
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\(P(WHITE) = \frac{0}{15} = 0\)
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\(P(BLUE) = \frac{4}{15}\)
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\(P(not\ RED) = P(BLUE) + P(GREEN) = \ \frac{4}{15} + \frac{5}{15} = \frac{9}{15} = \frac{3}{5}\)
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\(P(RED\ or\ WHITE) = P(RED) + P(WHITE) = \frac{6}{15} + 0 = \frac{6}{15}\)