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Question 2 (variant of question 1)
A digit is drawn from digits 0 to 9 (inclusive), followed by another draw of digit. What is the probability that the product of these 2 digits is an even number when
(i) Second digit cannot be same as first digit
(ii) Second digit can be same as first digit.
Too tough for me to handle these questions. Can someone please help me with them
Regarding your 2nd Question, here is my approach:
For (i) - Second digit cannot be same as first
Total number of ways to draw 2 digits one after the other = 10 x 9 = 90.
If first digit is even, then for product to be even, 2nd digit can be either even or odd.
Therefore, no. of possible ways = 4 (no. of even digits) x 8 (excluding 0 and the one already selected) = 32
If first digit is odd, then for product to be even, 2nd digit must be even.
no. of possible ways = 5 (no. of odd digits) x 4 (no. of even digits) = 20
Therefore, total ways = 32+20=52
Probabilty = 52/90 = 26/45
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For (ii) - Second digit can be same as first
Total number of ways to draw 2 digits one after the other = 10x10 = 100.
If first digit is even, then for product to be even, 2nd digit can be either even or odd.
Therefore, no. of possible ways = 4 (no. of even digits) x 9 (excluding 0) = 36.
If first digit is odd, then for product to be even, 2nd digit must be even.
no. of possible ways = 5 (no. of odd digits) x 4 (no. of even digits) = 20
Therefore, total ways = 36+20 = 56
Probabilty = 56/100 = 14/25
Hope these are correct answers...
Cheers!
Satish