
There are 4 white and 3 black balls in a box. In another box there are 3 white and 4 black balls. An unbiased dice is rolled. If it shows a number less than or equal to 3, then a ball is drawn from the first box, but if it shows a number more than 3, then a ball is drawn from the second box. If the ball drawn is black, then the probability that the ball was drawn from the first box is
A. \[\dfrac{1}{2}\]
B. \[\dfrac{6}{7}\]
C. \[\dfrac{4}{7}\]
D. \[\dfrac{3}{7}\]
Answer
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Hint: Find the probability of choosing the first and the second box and using those values try to find the probability of choosing a black ball, use conditional probability for finding the required value.
Complete step-by-step answer:
Let event A be the event of black ball chosen from the first box.
Let event B be the event that the chosen ball is black
Now we need to find \[P\left( {\dfrac{A}{B}} \right)\]
Now the event A is basically the ball chosen from firstbox \[ \cap \] chosen ball is black.
Probability of choosing the first box \[ = \dfrac{3}{6} = \dfrac{1}{2}\]
Probability of choosing the second box \[ = \dfrac{3}{6} = \dfrac{1}{2}\]
Now the Probability of choosing a black ball = (Probability of choosing the first bag \[ \times \] Probability of choosing black) \[ + \] (Probability of choosing the second bag \[ \times \] Probability of choosing black)
Now the probability of choosing a black ball from first box is \[\dfrac{3}{7}\] as there are total 7 balls and 3 of them are black, Similarly the probability of choosing a black ball from second box is \[\dfrac{4}{7}\]
So after putting all of them in the formula we will get
\[\begin{array}{l}
= \left( {\dfrac{1}{2} \times \dfrac{3}{7}} \right) + \left( {\dfrac{1}{2} \times \dfrac{4}{7}} \right)\\
= \dfrac{3}{{14}} + \dfrac{4}{{14}}\\
= \dfrac{{3 + 4}}{{14}}\\
= \dfrac{7}{{14}}\\
= \dfrac{1}{2}
\end{array}\]
Now,
\[\begin{array}{l}
P\left( {\dfrac{A}{B}} \right) = \dfrac{{\dfrac{1}{2} \times \dfrac{3}{7}}}{{\dfrac{1}{2}}}\\
P\left( {\dfrac{A}{B}} \right) = \dfrac{3}{7}
\end{array}\]
So, the correct answer is “Option D”.
Note: It must be noted that the probability of any outcome is given by the number of favourable possible outcomes divided by the total number of outcomes, and again there are certain other forms of probability like conditional probability and Bayes theorem. In this question we have used the normal form of probability and the conditional probability.
Complete step-by-step answer:
Let event A be the event of black ball chosen from the first box.
Let event B be the event that the chosen ball is black
Now we need to find \[P\left( {\dfrac{A}{B}} \right)\]
Now the event A is basically the ball chosen from firstbox \[ \cap \] chosen ball is black.
Probability of choosing the first box \[ = \dfrac{3}{6} = \dfrac{1}{2}\]
Probability of choosing the second box \[ = \dfrac{3}{6} = \dfrac{1}{2}\]
Now the Probability of choosing a black ball = (Probability of choosing the first bag \[ \times \] Probability of choosing black) \[ + \] (Probability of choosing the second bag \[ \times \] Probability of choosing black)
Now the probability of choosing a black ball from first box is \[\dfrac{3}{7}\] as there are total 7 balls and 3 of them are black, Similarly the probability of choosing a black ball from second box is \[\dfrac{4}{7}\]
So after putting all of them in the formula we will get
\[\begin{array}{l}
= \left( {\dfrac{1}{2} \times \dfrac{3}{7}} \right) + \left( {\dfrac{1}{2} \times \dfrac{4}{7}} \right)\\
= \dfrac{3}{{14}} + \dfrac{4}{{14}}\\
= \dfrac{{3 + 4}}{{14}}\\
= \dfrac{7}{{14}}\\
= \dfrac{1}{2}
\end{array}\]
Now,
\[\begin{array}{l}
P\left( {\dfrac{A}{B}} \right) = \dfrac{{\dfrac{1}{2} \times \dfrac{3}{7}}}{{\dfrac{1}{2}}}\\
P\left( {\dfrac{A}{B}} \right) = \dfrac{3}{7}
\end{array}\]
So, the correct answer is “Option D”.
Note: It must be noted that the probability of any outcome is given by the number of favourable possible outcomes divided by the total number of outcomes, and again there are certain other forms of probability like conditional probability and Bayes theorem. In this question we have used the normal form of probability and the conditional probability.
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