A basket contains 12 fruits: orange, apple and avocado pear, all of the same size. The number of oranges, apples and avocado pear forms three consecutive integers.
Two fruits are drawn one after the other without replacement. Calculate the probability that:
i. the first is an orange and the second is an avocado pear.
ii.both are of same fruit;
iii. at least one is an apple
They form 3 consecutive integers
let the number of orange be x, apple = x+1 and avocado pear = x+2
x + x+1 + x+2 = 12
3x + 3 = 12
3x = 9
x = 3
apple = 4
avocado pear = 5
probability that orange is drawn = \(\frac{3}{12}\)= \(\frac{1}{4}\)
probability that apple is drawn = \(\frac{4}{12}\)= \(\frac{1}{3}\)
probability that avocado pear is drawn = \(\frac{5}{12}\)
probability that the first is orange and the second is an avocado pear = \(\frac{3}{12}\) * \(\frac{5}{11}\) = \(\frac{5}{44}\)
ii.
both are of same fruit, they are both Orange or apple or Avocado pear
p(O∩O) + p(A∩A) + p(P∩P) = \(\frac{1}{4} * \frac{2}{11} + \frac{1}{3} * \frac{3}{11} + \frac{5}{12} * \frac{4}{11}\)
= \(\frac{1}{22} + \frac{1}{11} + \frac{5}{33}\)
= \(\frac{19}{66}\)
iii.
at least one is an apple = p(A∩O) or p(O∩A) or p(A∩P) or p(P∩A) or p(A∩A)
p(A∩O) + p(O∩A) + p(A∩P) + p(P∩A) + p(A∩A) =
= \(\frac{4}{12} * \frac{3}{11} + \frac{3}{11} * \frac{4}{12} + \frac{4}{12} * \frac{5}{11} + \frac{5}{12} * \frac{4}{11} + \frac{4}{12} * \frac{3}{11} \)
= \(\frac{12}{132}\) + \(\frac{12}{132}\) + \(\frac{20}{132}\) + \(\frac{20}{132}\) + \(\frac{12}{132}\)
= \(\frac{76}{132}\)
= \(\frac{19}{33}\)
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