Tag: baye's theorem

Questions Related to baye's theorem

A bag contains $(2n+1)$ coins. It is known that $n$ of these coins have a head on both sides, whereas the remaining $n+1$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\displaystyle \frac{31}{42}$, then $n$ is equal to 

business maths probability - iii baye's theorem bayes theorem probability and probability distribution

  1. $10$

  2. $11$

  3. $12$

  4. $13$

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Correct Option: A
Explanation:

Let ${ A } _{ 1 }$ denote the event that a coin having heads on both sides is chosen, and ${ A } _{ 2 }$ denote the vent that a fiar coin is chosen.
Let $E$ denote the vent that head occurs, Then
$\displaystyle P\left( { A } _{ 1 } \right) =\frac { n }{ 2n+1 } \Rightarrow P\left( { A } _{ 2 } \right) =\frac { n+1 }{ 2n+1 } $
Probability of occurrence of event $E$, if unfair coin was selected is $\displaystyle P\left( \frac { E }{ { A } _{ 1 } }  \right) =1$
Probability of occurrence of event $E$, if fair coin was selected is $\displaystyle P\left( \frac { E }{ { A } _{ 2 } }  \right) =\frac { 1 }{ 2 } $
$\because P\left( E \right) =P\left( { A } _{ 1 }\cap E \right) +P\left( { A } _{ 2 }\cap E \right) $
$\displaystyle \therefore P\left( E \right) =P\left( { A } _{ 1 } \right) P\left( \frac { E }{ { A } _{ 1 } }  \right) +P\left( { A } _{ 2 } \right) P\left( \frac { E }{ { A } _{ 2 } }  \right) $
$\displaystyle \Rightarrow \frac { 31 }{ 42 } =\frac { n }{ 2n+1 } .1+\frac { n+1 }{ 2n+1 } .\frac { 1 }{ 2 } \Rightarrow \frac { 31 }{ 42 } =\frac { 3n+1 }{ 2\left( 2n+1 \right)  } \ \Rightarrow 124n+62=126n+42\Rightarrow 2n=20\Rightarrow n=10$

The contents of urn I and II are as follows:
Urn I: 4 white and 5 black balls
Urn II: 3 white and 6 black balls
One urn is chosen at random and a ball is drawn and its colour is noted and replaced back to the urn. Again a ball is drawn from the same urn colour is noted and replaced. The process is repeated 4 times and as a result one ball of white colour and 3 of black colour are noted. Find the probability the chosen urn was I.

business maths probability - iii baye's theorem bayes theorem probability and probability distribution

  1. $\displaystyle \frac{125}{287}$

  2. $\displaystyle \frac{64}{127}$

  3. $\displaystyle \frac{25}{287}$

  4. $\displaystyle \frac{79}{192}$

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Correct Option: A

A signal which can be green or red with probability $\displaystyle \frac{4}{5}$ and $\displaystyle \frac{1}{5}$, respectively, is received at station A and then transmitted to station B. The probability of each station receiving the signal correctly is $\displaystyle \frac{3}{4}$. If the signal received at station B is green, then the probability that the original signal was green is

business maths probability - iii baye's theorem bayes theorem probability and probability distribution

  1. $\displaystyle \frac{3}{5}$

  2. $\displaystyle \frac{6}{7}$

  3. $\displaystyle \frac{20}{23}$

  4. $\displaystyle \frac{9}{20}$

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Correct Option: C
Explanation:
 Event $G$ = original signal is green
$E _1=A$ receives the signal correct
$E _2=B$ receives the signal correct
E = signal received by B is green
$P(\text{signal received by B is green}) = P(GE _1E _2)+ P(G\cap {E _1}\cap {E _2})+ P(\cap GE _1\cap{E _2})+ P(\cap G\cap {E _1}E _2)$
$P(E)=\dfrac {46}{5\times 16}$
$ P(G/E)=\dfrac {\dfrac {40}5\times 16}{\dfrac {46}5\times16}=\dfrac {20}{23}.$

One bag contains 3 white balls, 7 red balls and 15 black balls. Another bag contains 10 white balls, 6 red balls and 9 black balls. One ball is taken from each bag. What is the probability that both the balls will be of the same colour?

business maths probability - iii baye's theorem bayes theorem probability and probability distribution

  1. $207/625$

  2. $191/625$

  3. $23/625$

  4. $227/625$

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Correct Option: A
Explanation:
Bag $I=$ $3$ White $+$ ${7}$ Red $+$ $15$ Black

Bag $II=$ $10$ White $+$ ${6}$ Red $+$ $9$ Black

Each beg contains total of $25$ balls.

There are three cases for selection of a particular ball :

$1.$ White ball from Bag $I$ and Bag $II=\dfrac{3}{25}\times\dfrac{10}{25}$

$2.$ Red ball from Bag $I$ and Bag $II=\dfrac{7}{25}\times\dfrac{6}{25}$

$3.$ Black ball from Bag $I$ and Bag $II=\dfrac{15}{25}\times\dfrac{9}{25}$

$\therefore$ Total probability $=\dfrac{30}{625}+\dfrac{42}{625}+\dfrac{135}{25}=\dfrac{207}{625}.$

Hence, the answer is $\dfrac{207}{625}.$