Nine balls of the same size and colour, numbered 1, 2,…..9, were put into a packet. Now A draws a ball from packet, noted that it is of number a, and puts it back. Then B also draws a ball from the pocket and noted that it is of number b. Then probability for the inequality $a - 2 b + 10 > 0$ to hold is

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$\frac{60}{81}$

$\frac{61}{81}$

$\frac{52}{81}$

$\frac{59}{81}$

answer is D.

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Detailed Solution

Since each has equally 9 different possible results for A and B to draw a ball from the packet independently, the total number of possible events is $9^{2} = 81.$ From $a - 2 b + 10 > 0$ we get $2 b < a + 10.$ We find that when $b = 1, 2, 3, 4, 5$ a can take any value in $1, 2, 3, ....., 9$ to make the inequality hold. Then we have $9 \times 5 = 45$ admissible events
When b = 6, a can 3, 4, ….., 9 and there are 7 admissible events
When b = 7, a can 5, 6, 7, 8, 9 and there are 5 admissible events
When b = 8, a can 7, 8, 9 and there are 3 admissible events
When b = 9, a can 9 and there are 1 admissible events
So, the required probability is $\frac{45 + 7 + 5 + 3 + 1}{81} = \frac{61}{81}$