📢 Too many exams? Don’t know which one suits you best? Book Your Free Expert 👉 call Now!

    ⚡ Month End Offer- Few Hours Left Flat 55% Off On All Courses!! Coode - OCT55 Enroll Now ⚡
    00:00:00 AM Left

    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?
    A 2/75 Correct Answer Incorrect Answer
    B 1/75 Correct Answer Incorrect Answer
    C 1/150 Correct Answer Incorrect Answer
    D 7/150 Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    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 = ⁵P₃ × ⁵P₂ = 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

    Practice Next
    More Probability Questions

    Relevant for Exams:

    ask-question