Question

A bag contains 3 red and 7 black balls. Two balls are selected at random one by one
without replacement. What is the probability that one of them is black?
​

Answer 1

Answer:

Sample space is 10

Event of black ball are 7

Explanation:

Probability is 7/10

Answer 2

The probability that one of them is black ball " $\dfrac{2}{3}$".

Explanation:

• Assume that the P1 and P2 be the event that black ball is drawn in the first and second draw.

• Now we can draw the ball without replacement as:

        $B(P_1 \cap P_2) = B(\textrm{Both Black balls}) = \dfrac{7}{10} \times \dfrac{6}{9}\\\Rightarrow B= \dfrac{7}{15}$

• So that we can calculate the probability of black ball

       

      $B(P_2)= B(P_1 P_2) + B(P_2)\\\Rightarrow \dfrac{7}{10} \times \dfrac{6}{9} + \dfrac{3}{10} \times \dfrac{7}{9} = \dfrac{7}{15}+\dfrac{7}{30}\\\Rightarrow B(P_2)= \dfrac{14+7}{30} = \dfrac{7}{10}$

• Required probability for being a black ball is :

      $B= \dfrac{P_1 \Cap P_2}{B(P_2)}$

        $B = \dfrac{\frac{7}{15}} {\frac{3}{10}}\\B = \dfrac{7}{15} \times \dfrac{10}{7} \\\\B = \dfrac {2}{3}$