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**Hint:**In this question we are given two types of balls and we are given ways of drawing the balls. We need to find the probability that the third ball drawn is black. For this we will build four cases and work on them to find probabilities. Formula that we will use are:

(I) Probability of compound events $P\left( E \right)=P\left( {{E}_{1}} \right)\times P\left( \dfrac{E}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)P\left( \dfrac{E}{{{E}_{2}}} \right)+\cdots \cdots $.

(II) $P\left( \dfrac{A}{B} \right)$ refers to conditional probability where B event has already happened and we need to find probability of event A happening.

**Complete step-by-step solution:**

Here we are given total balls as (2+2) = 4. There are two white balls and two black balls. We need to find the probability of drawing the third ball as black. For this we will have possible cases to fill the ways in which order of balls can be picked. _____ ______ Black.

We will have four cases:

(1) ${{E}_{1}}:$ First balls could be white and other balls could also be white. Let us find the probability for that, for the first ball, we have a number of white balls as 2 and total balls as 4. So the probability of drawing the first ball as white becomes $\dfrac{2}{4}$.

We are given, if white ball is drawn we do not replace any ball. So we are left with 1 white ball and 3 total balls for the second draw. Probability of drawing the second ball as white becomes $\dfrac{1}{3}$.

Total probability for this case becomes $\dfrac{2}{4}\times \dfrac{1}{3}=\dfrac{2}{12}=\dfrac{1}{6}$.

$\therefore P\left( {{E}_{1}} \right)=\dfrac{1}{6}$

(2) ${{E}_{2}}:$ First ball could be white and second ball could be black. Let us find the probability for that. For the first ball, we have a number of white balls as 2 and total balls as 4. So the probability of drawing the first ball as white becomes $\dfrac{2}{4}$. Since no ball is replaced if the ball picked is white. So, we are left with 3 balls in total. Number of black balls is 2. Therefore, the probability of picking the second ball as black becomes $\dfrac{2}{3}$.

Total probability for this case become $\dfrac{2}{4}\times \dfrac{2}{3}=\dfrac{4}{12}=\dfrac{1}{3}$.

$\therefore P\left( {{E}_{2}} \right)=\dfrac{1}{3}$

(3) ${{E}_{3}}:$ First ball drawn could be black and other balls drawn could also be black. Let us find the probability for that. We have a number of black balls as 2 and a total number as 4. Therefore, the probability of picking the first ball as black becomes $\dfrac{2}{4}$.

We are given that, if black ball is drawn then we replace it with two balls of the same colour i.e. if we have picked black ball, we are left with 1 black ball and 3 total balls. But now we replaced it by adding two more black balls which makes the number of black balls as 3 and total balls as 5. So the probability of picking the second ball as black becomes $\dfrac{3}{5}$.

Total probability for this case becomes $\dfrac{2}{4}\times \dfrac{3}{5}=\dfrac{6}{20}=\dfrac{3}{10}$.

$\therefore P\left( {{E}_{3}} \right)=\dfrac{3}{10}$

(4) ${{E}_{4}}:$ First ball drawn could be black and other balls drawn could be white. Let us find the probability for that. We have a number of black balls as 2 and total balls as 4. So the probability of picking the first ball as black becomes $\dfrac{2}{4}$.

Since black ball is drawn so we have to replace it with two black balls which make total balls as 5. Number of white balls still remains 2. Therefore, the probability of picking the second ball as white becomes $\dfrac{2}{5}$.

Total probability for this case becomes $\dfrac{2}{4}\times \dfrac{2}{5}=\dfrac{4}{20}=\dfrac{1}{5}$.

$\therefore P\left( {{E}_{4}} \right)=\dfrac{1}{5}$

Now we need the third ball as black. So let us suppose E as the event of picking the third ball as black.

So now we will find conditional probabilities $P\left( \dfrac{E}{{{E}_{1}}} \right),P\left( \dfrac{E}{{{E}_{2}}} \right),P\left( \dfrac{E}{{{E}_{3}}} \right),P\left( \dfrac{E}{{{E}_{4}}} \right)$.

We know, $P\left( \dfrac{A}{B} \right)$ means that a B event has happened and we have to find the probability of event A happening.

(1) $P\left( \dfrac{E}{{{E}_{1}}} \right)$ will be the probability of drawing a third ball as black ball if two white balls are already drawn. So the number of balls left are 2 and the number of black balls left are also 2. So, $P\left( \dfrac{E}{{{E}_{1}}} \right)=\dfrac{2}{2}=1\therefore P\left( \dfrac{E}{{{E}_{1}}} \right)=1$.

(2) $P\left( \dfrac{E}{{{E}_{2}}} \right)$ will be probability of drawing third ball as black ball if one ball drawn is white and then second ball drawn is black.

After drawing white ball, we are left with 1 white ball in the urn. After the second drawn ball, we were left with 1 black ball but since it was black, we have to replace it with 2 black balls. So the number of black balls left becomes 3 balls. And thus total balls in the urn are 4. So $P\left( \dfrac{E}{{{E}_{2}}} \right)=\dfrac{3}{4}$.

(3) $P\left( \dfrac{E}{{{E}_{3}}} \right)$ will be the probability of drawing the third ball as black if both first and second ball drawn were black. After drawing one black ball, we replaced it with two more black balls, so we got 3 black balls after the first draw. After the second draw as black, we were left with 2 black balls but we again replaced it with 2 more balls. So there are 4 black balls after the second draw. Number of white balls are the same, so total balls for the third draw are 6 with 4 black balls. So we have $P\left( \dfrac{E}{{{E}_{3}}} \right)=\dfrac{4}{6}=\dfrac{2}{3}$.

(4) $P\left( \dfrac{E}{{{E}_{4}}} \right)$ will be the probability of drawing a third black ball if the first ball drawn was black and second ball drawn was white.

After the first ball drawn as black, we were left with one black ball but we replaced it with 2 more balls. So we get a total of 3 black balls. After the second draw, white ball was picked so we are left with 2 white balls. Total balls left are 3+1 = 4 balls. So, $P\left( \dfrac{E}{{{E}_{4}}} \right)=\dfrac{3}{4}$.

At last, we know that by theorem for compound events we have,

\[\begin{align}

& P\left( E \right)=P\left( {{E}_{1}} \right)\times P\left( \dfrac{E}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)P\left( \dfrac{E}{{{E}_{2}}} \right)+P\left( {{E}_{3}} \right)P\left( \dfrac{E}{{{E}_{3}}} \right)+P\left( {{E}_{4}} \right)P\left( \dfrac{E}{{{E}_{4}}} \right) \\

& \Rightarrow \dfrac{1}{6}\times 1+\dfrac{1}{3}\times \dfrac{3}{4}+\dfrac{1}{5}\times \dfrac{3}{4}+\dfrac{3}{10}\times \dfrac{2}{3} \\

& \Rightarrow \dfrac{1}{6}+\dfrac{1}{4}+\dfrac{3}{20}+\dfrac{1}{5} \\

& \Rightarrow \dfrac{10+15+9+12}{60} \\

& \Rightarrow \dfrac{46}{60} \\

& \Rightarrow \dfrac{23}{30} \\

\end{align}\]

**Hence the required probability is $\dfrac{23}{30}$.**

**Note:**Students should not forget to consider every possibility of solving this sum. They usually take three cases only thinking black white black same as for white black black. Keep in mind all the formulas and proceed carefully. Make sure that probability lies between 0 and 1.

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