What in the unconditional probability that the $2^{\text {rd }}$ ball drawn is Green?

$\text { i.e } P\left(G_2\right)$

Now the event $G_2=R_1 G_2 \quad G_1 G_2$

$P\left(G_2\right)=P\left(R_1 G_2 \cup G_1 G_2\right)$

$\therefore P\left(G_2\right) =P\left(R_1 \cap G_2\right)+P\left(G_1 \cap G_2\right) =P\left(G_2 \mid R_1\right) \cdot P\left(R_1\right)+P\left(G_2 \mid G_1\right) \cdot P\left(G_1\right)$

$\left.=\frac{8}{19} \times \frac{3}{5}+\frac{7}{19} \times \frac{2}{5}=\frac{24+14}{19 \times 5}=\frac{38}{19 \times 5}=\frac{2}{5}\right)$

What is the unconditional probablitiy that the $3^{rd}$ ball drawn is red?

Solution:

The 3rd ball drawn can red in union of four disjoint events.

$\ R_1 \cap R_2 \cap R_3 \cup R_1 \cap G_2 \cap R_3 \cup \cup G_1 \cap R_2 \cap R_3 \cup G_1 \cap G_2 \cap R_3$

$\therefore ~ ~ P\left(R_3\right)=P\left(R_1 R_2 R_3\right)+P\left(R_1 G_2 R_3\right) +P\left(G_1 R_2 R_3\right) +P\left(G_1 G_2 R_3\right) .$

$P\left(R_3\right)=\frac{3}{5} \times \frac{11}{19} \times \frac{10}{18}+\frac{3}{5} \times \frac{8}{19} \times \frac{4}{18}+\frac{2}{5}$

$\times \frac{2}{19} \times \frac{11}{18}+\frac{2}{5} \times \frac{7}{19} \times \frac{12}{18}$

$=\frac{1}{5 \times 19 \times 18}(330+264+264+168)$

$=\frac{102657}{5 \times 19 \times 18}=\frac{3}{5}$

Suppose you have 3 fair dice $\ni P(1)=P(2) =P(3)=\cdots=P(6)=\frac{1}{6}$

Also you have a fake dice which has four faces as $5$ & two faces as $6$

You choose 1 of the four dice at random and throw it. You get a 5

(a) What is the probability of getting a 5 ?

(b) Given that you have got a 5 what is the probability that you have chosen the fake die?

(a) $P(5)=P(5 / \text { fair die }) \times P(\text { fair die })$

$+P(5 / \text { fake die }) \times P(\text { fake die})$

$=\frac{1}{6} \times \frac{3}{4}+\frac{2}{3} \times \frac{1}{4}$

$=\frac{3}{24}+\frac{4}{24}=\frac{7}{24}$

(b) $P(\text { chosen the fake die } \mid \text { result is 5) } \longrightarrow$ bayes' theorem

$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(B \mid A) \cdot P(A)}{P(B)}$

$P(\text { fake die } \mid 5)$

$=\frac{P(5 \cap \text { fake die })}{P(5)} =\frac{P(5 \mid \text { fake die }) \times P(\text { fake die })}{P(5) \checkmark}$

$=\frac{\frac{2}{3} \times \frac{1}{4}}{\frac{7}{24}}=\frac{2}{12} \times \frac{24}{7}=\frac{4}{7}$

Ans.

(a) $\begin{aligned} & =\frac{1}{6} \times \frac{3}{4}+\frac{2}{3} \times \frac{1}{4} \ & =\frac{3}{24}+\frac{4}{24}=\frac{7}{24} \text {. } \ & \end{aligned}$

Bayes' Theorem:

$P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(B \mid A) \cdot P(A)}{P(B)}$

$P(\text { fake die } \mid 5)$

$=\frac{P(5 \cap \text { fake die })}{P(5)}$

$=\frac{P(5 \mid \text { fake die }) \times P(\text { fake die })}{P(5) \checkmark}$

$=\frac{\frac{2}{3} \times \frac{1}{4}}{\frac{7}{24}}=\frac{2}{12} \times \frac{24}{7}=\frac{4}{7}$

Let $E$ be the event that you choose the $5^{th}$ coin & $B$ be the event that you got a $H$

we want to compute $P(E \mid B)$.

Using Bayes' Theorem

$P(E \mid B)=\frac{P(B \cap E)}{P(B)}=\frac{P(B \mid E) \times P(E)}{P(B)}$

Now $P(B \mid E)=\frac{5}{10}=\frac{1}{2}$.

$2 P(E)=\frac{1}{10}$ as the coin is chosen randmoly.

$\therefore P(E \mid B)=\frac{\frac{1}{2} \times \frac{1}{10}}{P(B)}=P(\text { gating a } H)$

Now

$P(B)=P\left(1^{st} coin \cap B\right)+P\left(2^d \cap B\right) +\cdots+P\left(10^{\text {th }} \cap B\right)$

$=P(H \mid 1^{st} coin ). P(1^{st} coin) + \dots + P(H \mid 10^{th} coin). P(10^{th} coin)$

$=\frac{1}{10} \times \frac{1}{10}+\frac{2}{10} \times \frac{1}{10}+\cdots+\frac{10}{10} \times \frac{1}{10}$

$=\frac{1}{100}(1+2+\cdots+10)=\frac{1}{100} \times \frac{10 \times 11}{2}=\frac{11}{20}$

$\therefore P(E \mid B) =\frac{\frac{1}{20}}{\frac{11}{20}}=\frac{1}{11}$

Suppose a student is answering a question an MCQ question with 5 optionsof which only one is correct.

Now let P be the probability that the student gives the answer i.e. he randomly ticks and let 1-p be the probability that he knows the answer and therefore he ticks correctly.

Suppose the student ticks correctly. What is the probability that he gives the answer.

We want to compute P(guess/ ticked rightly)

Let E be the event that he guessed and let B be the event that he ticked it correctly.

Let $E$ be the event that he guessed it & $B$ be the event that he ticked it correctly

$\therefore P(E \mid B)=\frac{P(B \mid E) \cdot P(E)}{P(B)}$

Now $P(B \mid E)=\frac{1}{5},$ & $P(E)=p$ (given)

And

$P(B) =P(B \mid \text {guess}) \cdot P(\text {guess})+P(B \mid \text {know}) \cdot P(\text {know})$

$=\frac{1}{5} \times p+1 \cdot(1-p)=\frac{p+5(1-p)}{5}=\frac{5-4 P}{5}$

$\therefore P(E \mid B) =\frac{p}{5}$ \ $\frac{5-4 P}{5}=\frac{p}{5-4 P}$

Suppose you have 3 bags $A, B$ & $C$. The contents are as follows:

$A: 1W+2G+3 R$

$B: 2W+1 G+1 R$

$C: 4W+5G+3 R .$

You choose a bag at random & take out two balls from it. Suppose you get $1W$ & $1R$ ball. What is the probability that you have choose Bag A.?

Let $E$ be the event of selecting bag $A$.

$B$ be the event of getting $1W+1R$.

We want to compute $P(E \mid B)$

Using Bayers' Theorem: $P(E \mid B)=\frac{P(B \mid E) \cdot P(E)}{P(B)}$

Now $P(E)=\frac{1}{3}$ and $P(B \mid E)=\frac{3}{{}^6{c_2}}=\frac{3}{\frac{6!}{2!4!}}$

$=\frac{3}{\frac{5 \times 6}{2}}=\frac{1}{5} .$

Similary ,$P(1W+1 R \mid Bag B)^2=\frac{2}{{}^4c_2}=\frac{2}{\frac{4 !}{2 ! 2 !}}=\frac{2}{6}=\frac{1}{3}$

$P(1W+1 R \mid Bag C)=\frac{12}{12 c_2}=\frac{12}{\frac{11\times12}{2}}=\frac{2}{11}$

$\therefore P(B) =P(1W+1R)$

$=\frac{1}{5} \times \frac{1}{3}{}+\frac{1}{3} \times \frac{1}{3}+\frac{1}{3} \times \frac{2}{11}$

$=\frac{1}{3}\left(\frac{1}{5}+\frac{1}{3}+\frac{2}{11}\right)=\frac{1}{3}\left(\frac{33+55+30}{165}\right)$

$=\frac{1}{3} \times \frac{118}{165}$

$\therefore P(E \mid B) =\frac{\frac{1}{5} \times \frac{1}{3}}{\frac{1}{3} \times \frac{118}{165}}=\frac{1}{5} \times \frac{165}{118}=\frac{33}{118}$

Suppose yon are a waiting for a letter from KOLKATA, KATHMANDU and MUSCAT.

Suppose you receive a letter and the only thing legible in the address is two successive letters: AT

What in the probability that it is from KOLKATA?

$P\left(K O L K A T A \mid AT\right) =\frac{P\left(AT \mid \text { KOLKATA }\right) \times P(K O L K A T A)}{P\left(AT\right)}$

Now

$P(A T \mid \text { KOLKATA) } : \frac{1}{6}$

$P(A T \mid \text { KATHMANDU)} : \frac{1}{8}$

$P(A T \mid \text { MUSCAT) }=\frac{1}{5}$

$P(K O L K A T A \mid A T) =\frac{\frac{1}{6} \times \frac{1}{3}}{\frac{1}{6} \times \frac{1}{3}+\frac{1}{8} \times \frac{1}{3}+\frac{1}{5} \times \frac{1}{3}}$

$=\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{8}+\frac{1}{5}} =\frac{\frac{1}{6}}{\frac{40+30+48}{5 \times 6 \times 8}}$

$=\frac{40}{40+30+48}=\frac{40}{118}=\frac{20}{59}$

Suppose India & Australia are playing a five -match test series. Suppose Virt kohli won the toss in the first four matches. What is the probability that he will win the toss in the fifth match as well?

(A) $1 \times$ $~ ~$ (B) $\frac{1}{5}$ $~ ~$(c) 0 $~ ~$ (D) $\frac{1}{2}$

Solution:

We assume that the toss is done using an unbiased coin, so that in each toss the probabilist of winning is equal for both the captains.

Can it be 1? Because since viral won all the previous tosses, therefore he will win the fifth toss.

Can it be $\frac{1}{5}$ ? Because Virat won all the 4 tosses therefore now the Australian captain will have $\frac{4}{5}$ probability of winning.

Can it be 0 ? Because now Australian captain $X$ will have to win the toss

$\therefore ~ P \text { (virat winning) }=0$

Correct option is (D) i.e. Answer is $\frac{1}{2}$

$P$ (virat winning $5^{\text {th }}$ toss $\mid$ he won all the previus tosses $)$ is same as $P$ (virat winning the toss) as there are independent events.