Nine balls of the same size and colour, numbered , were put into an Urn. Now A draws a ball from Urn, noted that it is of number , and puts it back. Then B also drawn a ball from the Urn and noted that it is of number . Then probability that the inequalit

Question

Nine balls of the same size and colour, numbered $1, 2, \dots, 9$ , were put into an Urn. Now A draws a ball from Urn, noted that it is of number $x$ , and puts it back. Then B also drawn a ball from the Urn and noted that it is of number $y$ . Then probability that the inequality $x - 2 y + 10 > 0$ to hold is

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(4)

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 from 1, 2, 3, ..., 9 to make the inequality hold. Then we have 9 × 5 = 45 admissible events.

When b = 6, a can be 3, 4, ..., 9 and there are 7 admissible events.

When b = 7, a can be 5, 6, 7, 8, 9 and there are 5 admissible events.

When b = 8, a can be 7, 8, 9 and there are 3 admissible events.

When b = 9, a can be 9 and there is 1 admissible event.

So, the required probability is $\frac{45 + 7 + 5 + 3 + 1}{81} = \frac{61}{81}$

Solution

1 $\frac{52}{18}$

2 $\frac{59}{81}$

3 $\frac{60}{81}$

4 $\frac{61}{81}$

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Solution

Q. Nine balls of the same size and colour, numbered 1,2,…,9, were put into an Urn. Now A draws a ball from Urn, noted that it is of number x, and puts it back. Then B also drawn a ball from the Urn and noted that it is of number y. Then probability that the inequality x-2y+10>0 to hold is

1 5218

2 5981

3 6081

4 6181

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1 $\frac{52}{18}$

2 $\frac{59}{81}$

3 $\frac{60}{81}$

4 $\frac{61}{81}$