Question

In a bag there are six balls of unknown colors; three balls are drawn one by one without replacement and found to be black; find the probability that no black ball is left in the bag.

Answer 1

Hint: In this particular type of question use the concept that balls are drawn one by one without replacement so after drawing one ball total number of balls remaining in the bag is one less so probability of next drawn changed accordingly and the probability of drawing three black balls one by one is the multiplication of the probabilities of drawing black ball one by one and construct all the possible cases so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data-
There are 6 balls of unknown colors in the bag.
Now three balls are drawn one by one and found to be black without replacement.
So we have to find the probability that there are one three balls in the bag.
Now as we know that the probability is the ratio of favorable number of outcomes to the total number of outcomes.
$ \Rightarrow P = \dfrac{{{\text{favorable number of outcomes}}}}{{{\text{total number of outcomes}}}}$
Now construct the possible cases:
$\left( i \right)$ If there are 6 black balls in the bag
So the probability of drawing one black ball = (6/6), so the remaining black ball in the bag = (6 – 1) = 5, now the probability of drawing second ball from the bag = (5/5), so the remaining black ball in the bag = (5 – 1) = 4, now the probability of drawing third ball from the bag = (4/4).
So the probability of drawing three black balls one by one is the multiplication of the above evaluated probability.
So the probability of drawing 3 black balls one by one without replacement = $\dfrac{6}{6}.\dfrac{5}{5}.\dfrac{4}{4} = 1$
$\left( {ii} \right)$ If there are five black ball in the bag
So the probability of drawing 3 black balls one by one without replacement = $\dfrac{5}{6}.\dfrac{4}{5}.\dfrac{3}{4} = \dfrac{1}{2}$
$\left( {iii} \right)$ If there are four black ball in the bag
So the probability of drawing 3 black balls one by one without replacement = $\dfrac{4}{6}.\dfrac{3}{5}.\dfrac{2}{4} = \dfrac{1}{5}$
$\left( {iv} \right)$ If there are three black ball in the bag
So the probability of drawing 3 black balls one by one without replacement = $\dfrac{3}{6}.\dfrac{2}{5}.\dfrac{1}{4} = \dfrac{1}{{20}}$
So the probability that there are three balls in the bag is the ratio of probability if there are three black balls in the bag to the addition of probabilities of all the above cases.
So the probability that there are one three balls in the bag $ = \dfrac{{\dfrac{1}{{20}}}}{{1 + \dfrac{1}{2} + \dfrac{1}{5} + \dfrac{1}{{20}}}}$
Now simplify we have,
So the probability that there are one three balls in the bag $ = \dfrac{{\dfrac{1}{{20}}}}{{\dfrac{{20 + 10 + 4 + 1}}{{20}}}} = \dfrac{1}{{35}}$
So this is the required answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall the probability formula which is stated above then construct all the possible cases according to given condition as above and the probability that there are three balls in the bag is the ratio of probability if there are three black ball in the bag to the addition of probabilities of the all the construct cases.