Probability Test - 62

Let us now assume, \(\mathrm{X}\) represents the number of correct answers by guessing in the multiple-choice set

Now, probability of getting a correct answer, \(p=\frac{1 }{3}\)

Thus, \(q=1-p=1-\frac{1}{3}\)

\(=\frac{2}{3}\)

Clearly, we have \(\mathrm{X}\) is a binomial distribution where \(\mathrm{n}=5\) and \(\mathrm{P}=\frac{1}{3}\)

\(P(X=x)={ }^{n} C_{x} q^{n-x} p^{x} \)

\(={ }^{5} C_{x}\left(\frac{2}{3}\right)^{5^{-x}} \cdot\left(\frac{1}{3}\right)^{x}\)

Hence, probability of guessing more than 4 correct answer \(=P(X \geq 4)\)

\(=P(X=4)+P(X=5)\)

\(={ }^{5} C_{4}\left(\frac{2}{3}\right) \cdot\left(\frac{1}{3}\right)^{4}+{ }^{5} C_{5}\left(\frac{1}{3}\right)^{5}\)

\(=5 \cdot (\frac{2}{3}) .(\frac{1}{81})+1 \cdot (\frac{1}{243})\)

\(=\frac{11}{243}\)

Let \(E_{1}\) be the event of time consumed by machine \(A, E_{2}\) be the event of time consumed by machine \(B\) and \(E_{2}\) be the event of time consumed by machine \(C\). Let \(X\) be the event of producing defective items.

Then \(P\left(E_{1}\right)=50 \%=\frac{50}{100}\)

\(=\frac{1}{2}\)

\(P\left(E_{2}\right)=30 \%=\frac{30}{100}\)

\(=\frac{3}{10}\)

\(P\left(E_{3}\right)=20 \%_{b}=\frac{20}{100}\)

\(=\frac{1}{5}\)

As we a headed coin has head on both sides so it will shows head.

Also \(P\left(\frac{X }{E_{1}}\right)=P(\) defective item produced by \(A)=1 \%\)

\(=\frac{1}{100}\)

And \(P\left(\frac{X}{ E_{2}}\right)=P\) (defective item produced by \(\left.B\right)=5 \%\)

\(=\frac{5}{100}\)

And \(P\left(\frac{X}{ E_{3}}\right)=P\) (defective item produced by \(\left.C\right)=7 \%\)

\(=\frac{7}{100}\)

Now the probability that item produced by machine \(A\), being given that defective item is produced, is \(P\left(\frac{E_{1}}{ A}\right)\).

By using Bayes' theorem, we have

\(\mathrm{P}\left(\frac{\mathrm{E}_{1}}{ \mathrm{X}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{X}}{\mathrm{E}_{1}}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{X}}{\mathrm{E}_{1}}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{X}}{\mathrm{E}_{2}}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \cdot \mathrm{P}\left(\frac{\mathrm{X}}{\mathrm{E}_{3}}\right)}\)

Now by substituting the values we get

\(=\frac{(\frac{1}{2}) \cdot (\frac{1}{100})}{(\frac{1}{2}) \cdot (\frac{1}{100})+(\frac{3}{10}) \cdot (\frac{5}{100})+(\frac{1}{5}) \cdot (\frac{7}{100})}\)

\(=\frac{(\frac{1}{2} )\cdot (\frac{1}{100})}{(\frac{1}{100})\left((\frac{1}{2})+(\frac{3}{2})+(\frac{7}{5})\right)}\)

\(=\frac{(\frac{1}{2})}{(\frac{17}{5})}=(\frac{5}{34} )\)

\(\Rightarrow \mathrm{P}\left(\frac{\mathrm{E}_{1}}{\mathrm{X}}\right)=\frac{5}{34}\)

Given,\(P(A)=2 P(B)=6 P(C)\)

We know that when events are mutually exclusive and exhaustive,

\(P(A)+P(B)+P(C)=1\)

Let, \(P(A)\) = \(k\)

Then, \(P(B)= \frac{P(A)}{2}=\frac{k}{2}\)

\(P(B)=\frac{k}{2}\)

\(\Rightarrow P(C)= \frac{k}{6}\)

So, according to the concept\(k+\frac{k}{2}+\frac{k}{6}=1\)

\(\Rightarrow \frac{10 k}{6}=1\)

\(\Rightarrow k=\frac{3}{5}\)

Therefore,\(P(B)\) = \(\frac{k}{2}=\frac{3}{5 \times 2}\)\(=0.3\)

Let \(E_{1}\) be the event that first group wins the competition, \(E_{2}\) be the event that that second group wins the competition and \(\mathrm{A}\) be the event of introducing a new product.

Then \(P\left(E_{1}\right)=0.6\) and \(P\left(E_{2}\right)=0.4\)

Also \(P\left(\frac{A}{ E_{1}}\right)=P\) (introducing a new product given that first group wins) \(=0.7\)

And \(P\left(\frac{A}{E_{2}}\right)=P\) (introducing a new product given that second group wins) \(=0.3\)

Now the probability of that new product introduced was by the second group, being given that a new product was introduced, is \(P\left(\frac{E_{2}}{A}\right)\).

By using Bayes' theorem, we have

\(\mathrm{P}\left(\frac{E_{2}}{ \mathrm{A}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{A}}{ \mathrm{E}_{1}}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)}\)

Now by substituting the values we get

\(=\frac{0.4 \times 0.3}{0.6 \times 0.7+0.4 \times 0.3}\)

\(=\frac{0.12}{0.42+0.12}\)

\(=\frac{0.12}{0.54}\)

\(=\frac{12}{54}\)

\(=\frac{2}{9}\)

\(\mathrm{P}\left(\frac{\mathrm{E}_{2}}{\mathrm{A}}\right)=\frac{2}{9}\)

Given,

A box of oranges.

Let \(A , B\) and \(C\) denotes respectively the events that the first, second and third drawn orange is good.

Now,

\(P(A)=P\) (good orange in first draw) \(=\frac{12}{15}\)

Because the second orange is drawn without replacement so, now the total number of good oranges will be 11 and the total oranges will be 14 that is the conditional probability of B given that A has already occurred.

Now,

\(P(\frac{B}{A})=P\) (good orange in second draw) \(=\frac{11}{14}\)

Because the third orange is drawn without replacement so, now the total number of good oranges will be 10 and total orangs will be 13 that is the conditional probability of \(C\) given that \(A\) and \(B\) has already occurred.

Now,

\(P(\frac{C}{AB})=P\) (good orange in third draw) \(=\frac{10}{13}\)

Thus the probability that all the oranges are good

= \(P(A \cap B \cap C)\) =\(\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13} \)

\(= \frac{44}{91}\)

\(\therefore\) The probability that a box will be approved for sale is \( \frac{44}{91}\).

Given,

An urn contains 5 red and 5 black balls.

Let in the first attempt the ball drawn is of red colour.

P (probability of drawing a red ball) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

Now the two balls of same colour (red) are added to the urn then the urn contains 7 red and 5 black balls.

P (probability of drawing a red ball) = \(\frac{7}{12}\)

Now,

Let in the first attempt the ball is drawn is of black colour.

P (probability of drawing a black ball) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

Now the two balls of the same colour (black) are added to the urn then the urn contains 5 red and 7 black balls.

P (probability of drawing a red ball) = \(\frac{5}{12}\)

The probability of drawing the second ball as of red colour is\(=\left(\frac{1}{2} \times \frac{7}{12}\right)+\left(\frac{1}{2} \times \frac{5}{12}\right)\)

\(=\frac{1}{2}\left(\frac{7}{12}+\frac{5}{12}\right)=\frac{1}{2} \times 1\)

\(=\frac{1}{2}\)

\(\therefore\) The probability of drawing the second ball as of red colour is\(\frac{1}{2}\).

Let \(E_{1}\) be the event that the outcome on the die is 5 or \(6, E_{2}\) be the event that the outcome on the die is \(1,2,3\) or 4 and \(\mathrm{A}\) be the event getting exactly head.

Then \(P\left(E_{1}\right)=\frac{2}{6}\)

\(=\frac{1}{ 3}\)

\(P\left(E_{2}\right)=\frac{4}{6}\)

\(=\frac{2}{3}\)

As in throwing a coin three times we get 8 possibilities.

(HHH, HHT, HTH, THH, TTH, THT, HTT, TTT)

\(\Rightarrow P\left(\frac{A}{ E_{1}}\right)=P\) (obtaining exactly one head by tossing the coin three times if she get 5 or 6\()=\frac{3}{8}\)

And \(P\left(\frac{A}{E_{2}}\right)=P\) (obtaining exactly one head by tossing the coin three times if she get \(1,2,3\) or 4\()=\frac{1 }{2}\)

Now the probability that the girl threw \(1,2,3\) or 4 with a die, being given that she obtained exactly one head, is \(\mathrm{P}\) \(\left(\frac{\mathrm{E}_{2}}{\mathrm{A}}\right)\)

By using Bayes' theorem, we have

\(\mathrm{P}\left(\frac{\mathrm{E}_{2}}{ \mathrm{A}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{1}}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)}\)

Now by substituting the values we get

\(=\frac{\frac{2}{3} \cdot \frac{1}{2}}{\frac{1}{3} \cdot \frac{3}{8}+\frac{2}{3} \cdot \frac{1}{2}}\)

\(=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}} \)

\(=\frac{\frac{1}{3}}{\frac{3+8}{24}}=\frac{8}{11}\)

\(\Rightarrow \mathrm{P}\left(\frac{\mathrm{E}_{2}}{\mathrm{A}}\right)=\frac{8}{11}\)

Given,

\(P(A)= 60\%\) = \(\frac{60}{100}\)

\(= 0.60\)

\(P(A') = 1 -0.60\)

\(= 0.40\)

\(P(\frac{B }{A})= 10\%\) = \(\frac{10}{100}\)

\(= 0.10\)

\(P(\frac{B}{A})^{\prime}=80\%\) = \(\frac{80}{100}\)

\(= 0.80\)

We know that,

\(P(\frac{A}{B}) \cdot P(B)=P(\frac{B}{A})-P(A)\)

Let, \(A\) be the event : the employees are graduate

Let, \(B\) be the event : the employees are in sales

\(P(B)=P(A) \cdot P(\frac{B }{ A})+P\left(A^{\prime}\right) \cdot P(\frac{B}{A})^{\prime}\)

\(\Rightarrow P(B) = 0.60 \times 0.10+0.40 \times 0.80\)

\(\Rightarrow P(B) = 0.38\)

\(\therefore\) The probability that an employee selected at random is in sales, is \(0.38\).

Let us assume, \(X\) denotes the events having a person heart attack

\({A}_{1}\) denote events having the selected person followed the course of yoga and meditation

And, \({A}_{2}\) denote the events having the person adopted the drug prescription

It is given in the question that,

\(P(X)=0.40\)

And, \(P\left(A_{1}\right)=P\left(A_{2}\right)=\frac{1}{2}\)

\(P\left(\frac{X}{A_{1}}\right)=0.40 \times 0.70=0.28\)

\(P\left(\frac{X}{ A_{2}}\right)=0.40 \times 0.75=0.30\)

Therefore,

Probability (The patient suffering from a heart attack and followed a course of meditation and yoga):

\(P\left(\frac{A_{1}}{ X}\right)=\frac{P\left(A_{1}\right) P\left(\frac{X }{ A_{1}}\right)}{P\left(A_{1}\right) P\left(\frac{X}{A_{1}}\right)+P\left(A_{2}\right) P\left(\frac{X}{A_{2}}\right)}\)

\(=\frac{\frac{1}{2} \times 0.28}{\frac{1}{2} \times 0.28+\frac{1}{2} \times 0.30} \)

\(=\frac{14}{29}\)

Given,

A lot of 30 bulbs which include 6 defectives.

Then number of non-defective bulbs \(=30-6\)

\(=24\)

As 4 bulbs are drawn at random with replacement.

Let \(X\) denotes the number of defective bulbs from the selected bulbs.

Clearly, \(X\) can take the value of \(0,1,2,3\) or 4 .

\(P(X=0)=P(4\) are non-defective and 0 defective)

\(=4_{C_{0}} \cdot \frac{4}{5}+\frac{4}{5} \cdot \frac{4}{5}.\frac{4}{5}\)

\(=\frac{256}{625}\)

\(P(X=1)=P(3\) are non-defective and 1 defective)

\(=4_{C_{1}} \cdot \frac{1}{5} \cdot\left(\frac{4}{5}\right)^{2}\)

\(=\frac{256}{625}\)

\(P(x=2)=P(2\) are non-defective and 2 defective)

\(=4_{C_{2}} \cdot\left(\frac{1}{5}\right)^{2} \cdot\left(\frac{1}{5}\right)^{2}\)

\(=\frac{96}{625}\)

\(P(X=3)=P(1\) are non-defective and 3 defective)

\(=4_{C_{3}}+\left(\frac{1}{5}\right)^{3}+\frac{4}{5}=\frac{16}{625}\)

\(P(X=4)=P(0\) are non-defective and 4 defective)

\(=4_{C_{4}}.\left(\frac{1}{5}\right)^{4}\)

\(=\frac{1}{625}\)