JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 2)
A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is :
$$\frac{3}{7}$$
$$\frac{5}{6}$$
$$\frac{5}{7}$$
$$\frac{2}{7}$$
Explanation
Let $E_{i} \rightarrow$ Bag have at least $i$ black balls
$E \rightarrow 2$ balls are drawn & both black
$$ \begin{aligned} & \therefore P\left(\frac{E_{5} \text { or } E_{6}}{E}\right)=\frac{P\left(\frac{E}{E_{5}}\right)+P\left(\frac{E}{E_{6}}\right)}{\sum\limits_{i=1}^{6} P\left(\frac{E}{E_{i}}\right)} \\\\ & =\frac{\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{6} C_{2}}{{ }^{6} C_{2}}}{0+\frac{{ }^{2} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{3} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{4} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{6} C_{2}}{{ }^{6} C_{2}}} \\\\ & =\frac{10+15}{1+3+6+10+15}=\frac{25}{35}=\frac{5}{7} \end{aligned} $$
$E \rightarrow 2$ balls are drawn & both black
$$ \begin{aligned} & \therefore P\left(\frac{E_{5} \text { or } E_{6}}{E}\right)=\frac{P\left(\frac{E}{E_{5}}\right)+P\left(\frac{E}{E_{6}}\right)}{\sum\limits_{i=1}^{6} P\left(\frac{E}{E_{i}}\right)} \\\\ & =\frac{\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{6} C_{2}}{{ }^{6} C_{2}}}{0+\frac{{ }^{2} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{3} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{4} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{6} C_{2}}{{ }^{6} C_{2}}} \\\\ & =\frac{10+15}{1+3+6+10+15}=\frac{25}{35}=\frac{5}{7} \end{aligned} $$
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