Question

Nine identical balls are numbered 1, 2 ,....9, are put in a bag. \[A\] draws a ball and gets the number \[a\]. The ball is put back in the bag. Next \[B\] draws a ball and gets the number \[b\]. The probability that \[a\] and \[b\] satisfies the inequality \[a - 2b + 10 > 0\] is
A. \[\dfrac{{52}}{{81}}\]
B. \[\dfrac{{55}}{{81}}\]
C. \[\dfrac{{61}}{{81}}\]
D. \[\dfrac{{62}}{{81}}\]

Answer 1

Hint:
Here we have to find the probability that the given variable satisfies the given inequality. For that, we will first find the possible ways of choosing the balls and then the total number of ways of choosing the balls. Then we will find the required probability which will be equal to the ratio of the possible ways of choosing the balls to the total number of ways of choosing the balls.

Complete step by step solution:
It is given that there are nine identical balls in a bag. Also it is given that when \[A\] draws a ball, he gets the number \[a\] and when \[B\] draws a ball, he gets the number \[b\].
The given inequality is
 \[a - 2b + 10 > 0\]
We can write \[b\] as
\[ \Rightarrow b < \dfrac{{a + 10}}{2}\]
We know that the minimum value of number \[b\] is 1.
Therefore,
\[1 \le b < \dfrac{{a + 10}}{2}\]
Now, we will substitute the value of \[a\] one by one from 1 to 9.
When \[a = 1\]
\[\begin{array}{l}1 \le b < \dfrac{{1 + 10}}{2}\\ \Rightarrow 1 \le b < \dfrac{{11}}{2}\end{array}\]
We know that the value of \[b\] can only be in whole numbers.
Therefore, we get
When \[a = 1\], then \[b = 1,2,3,4,5\] …………. \[\left( 1 \right)\]
When \[a = 2\]
\[\begin{array}{l}1 \le b < \dfrac{{2 + 10}}{2}\\ \Rightarrow 1 \le b < \dfrac{{12}}{2}\\ \Rightarrow 1 \le b < 6\end{array}\]
We know that the value of \[b\] can only be in whole numbers.
Therefore, we get
When \[a = 2\], then \[b = 1,2,3,4,5\] ……….. \[\left( 2 \right)\]
When \[a = 3\]
\[\begin{array}{l}1 \le b < \dfrac{{3 + 10}}{2}\\ \Rightarrow 1 \le b < \dfrac{{13}}{2}\end{array}\]
We know that the value of \[b\] can only be in whole numbers.
Therefore, we get
When \[a = 3\], then \[b = 1,2,3,4,5,6\] ................ \[\left( 3 \right)\]
When \[a = 4\]
\[\begin{array}{l} \Rightarrow 1 \le b < \dfrac{{4 + 10}}{2}\\ \Rightarrow 1 \le b < \dfrac{{14}}{2}\\ \Rightarrow 1 \le b < 7\end{array}\]
We know that the value of \[b\] can only be in whole numbers.
Therefore, we get
When \[a = 4\], then \[b = 1,2,3,4,5,6\] ................ \[\left( 4 \right)\]
When \[a = 5\]
\[\begin{array}{l}1 \le b < \dfrac{{5 + 10}}{2}\\ \Rightarrow 1 \le b < \dfrac{{15}}{2}\end{array}\]
We know that the value of \[b\] can only be in whole numbers.
Therefore, we get
When \[a = 5\], then \[b = 1,2,3,4,5,6,7\] ................ \[\left( 5 \right)\]
When \[a = 6\]
\[\begin{array}{l}1 \le b < \dfrac{{6 + 10}}{2}\\ \Rightarrow 1 \le b < \dfrac{{16}}{2}\\ \Rightarrow 1 \le b < 8\end{array}\]
We know that the value of \[b\] can only be in whole numbers.
Therefore, we get
When \[a = 6\], then \[b = 1,2,3,4,5,6,7\] ................ \[\left( 6 \right)\]
When \[a = 7\]
\[\begin{array}{l}1 \le b < \dfrac{{7 + 10}}{2}\\ \Rightarrow 1 \le b < \dfrac{{17}}{2}\end{array}\]
We know that the value of \[b\] can only be in whole numbers.
Therefore, we get
When \[a = 7\], then \[b = 1,2,3,4,5,6,7,8\] ................ \[\left( 7 \right)\]
When \[a = 8\]
\[\begin{array}{l}1 \le b < \dfrac{{8 + 10}}{2}\\ \Rightarrow 1 \le b < \dfrac{{18}}{2}\\ \Rightarrow 1 \le b < 9\end{array}\]
We know that the value of \[b\] can only be in whole numbers.
Therefore, we get
When \[a = 8\], then \[b = 1,2,3,4,5,6,7,8\] ................ \[\left( 8 \right)\]
When \[a = 9\]
\[\begin{array}{l}1 \le b < \dfrac{{9 + 10}}{2}\\ \Rightarrow 1 \le b < \dfrac{{19}}{2}\end{array}\]
We know that the value of \[b\] can only be in whole numbers.
Therefore, we get
When \[a = 9\], then \[b = 1,2,3,4,5,6,7,8,9\] ................ \[\left( 9 \right)\]
Therefore, number of possible pairs of the number \[a\] and \[b\] \[ = 5 + 5 + 6 + 6 + 7 + 7 + 8 + 8 + 9 = 61\]
Total number of ways of choosing the pairs \[ = 9 \times 9 = 81\]
Now, we will find the required probability which will be equal to the ratio of the possible ways of choosing the balls to the total number of ways of choosing the balls.
\[{\rm{Probability}} = \dfrac{{61}}{{81}}\]

Therefore, the correct option is option C.

Note:
Here we have solved the inequality to get the required pairs of the numbers. If it is written for a variable \[x\] that \[x < 5\], then this means that the possible values of the variable \[x\] are 1, 2, 3 and 4. However, the value of the variable \[x\] can’t be 5 but if it is written for a variable \[x\] that \[x \le 5\], then this means that the possible values of the variable \[x\] are 1, 2, 3, 4 and 5.