Question
A five digits number is chosen at random. What is the
probability that all the digits are distinct, the digits at odd places are odd and the digits are at even places are even?Solution
Odd digits – 1, 3, 5, 7, 9 Even digits – 0, 2, 4, 6, 8 Since, odd digits at odd places and even digits at even place. Places of odd digits – 3 Places of even digits – 2 ∴ Favourable ways = āµPā × āµPā = 5 × 4 × 3 × 5 × 4 = 1200 ∴ Total five digits numbers that can be formed = 9 × 10 × 10 × 10 × 10 = 90,000 Required probability = 1200/90000 = 1/75
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