Probability Test 24

Given: probability that electronic component fails when first used $$\displaystyle =0.10,$$ i.e., $$\displaystyle P\left( F \right) =0.10$$

Given,The probability of a Jee Aspirant to be successful if he studies for 10 hours per day,$$P(t_{10})=0.8$$ 
The probability of a Jee Aspirant to be successful if he studies for 7 hours per day,$$P(t_{7})=0.6$$
The probability of a Jee Aspirant to be successful if he studies for 4 hours per day,$$P(t_{4})=0.4$$
The probability that she will study 10 hours per day,$$\displaystyle P(A)=0.1$$
The probability that she will study 7 hours per day,$$\displaystyle P (B)=0.2$$
The probability that she will study 4 hours per day,$$\displaystyle P(C)=0.7$$
$$\therefore The\;probability\;that\;she\;will\;be\;successful,P(S)=P(t_{10})*P(A)+P(t_{7})*P(B)+P(t_{4})*P(S/t_{4})$$
$$=0.8*0.1+0.6*0.2+0.4*0.7=0.48$$
$$\therefore$$ The probability that she will not success,$$P(\overline S)=1-P(S)=1-0.48=0.52$$
But probability that she studies 4 hours per day but didn't success=$$P(\overline t_{4} \cap C)=P(\overline t_{4})P(C)=(1-0.4)*0.7=0.42$$
Since $$(A,t_{10}),(B,t_{7}),(C,t_{4})$$ pairwise independent $$\Rightarrow P(A \cap B)=P(A).P(B)$$
If $$(A,t_{10}),(B,t_{7}),(C,t_{4})$$ are independent,then $$(A,\overline t_{10}),(B,\overline t_{7}),(C,\overline t_{4}),(\overline A,t_{10}),(\overline B,t_{7}),(\overline C,t_{4})$$ will also be $$independent.$$
$$\therefore$$ The probability that she studies 4 hours per day given she is not successful,$$P(C/\overline S)=\displaystyle\frac{P(C \cap \overline S)}{P(\overline S)}$$
$$=\displaystyle\frac{0.42}{0.52}=\displaystyle\frac{21}{26}$$

Let the event of person goes to office by a car, scooter, bus or train be A, B, C and D, respectively.

We, have $$P(A)=\dfrac{1}{7},P(B)=\dfrac{3}{7}, P(C)=\dfrac{2}{7}$$ and $$P(D)=\dfrac{1}{7}$$

Let $$E=$$ He reached office in time

We have,

$$P\left(\dfrac{\overline{E}}{A}\right)=\dfrac{2}{9}, P\left(\dfrac{\overline{E}}{B}\right)=\dfrac{1}{9}, P\left(\dfrac{\overline{E}}{C}\right)=\dfrac{4}{9}$$ and $$ P\left(\dfrac{\overline{E}}{D}\right)=\dfrac{1}{9}$$

$$\therefore \displaystyle P\left(\frac{A}{E}\right) = \frac{P(A)\cdot P\left(\frac{E}{A}\right)}{P(A)\cdot P\left(\frac{E}{A}\right)+P(B)\cdot \left(\frac{E}{B}\right)}+P(C)\cdot P\left(\frac{E}{C}\right)+P(D)\cdot \left(\frac{E}{D}\right)$$

$$\displaystyle =\dfrac{\dfrac{1}{7}\cdot \dfrac{7}{9}}{\dfrac{1}{7}\cdot \dfrac{7}{9}+\dfrac{3}{9}\cdot \dfrac{8}{9}+\dfrac{2}{7}\cdot \dfrac{5}{9}+\dfrac{1}{7}\cdot \dfrac{8}{9}}$$

$$=\dfrac{7}{7+24+10+8}=\dfrac{7}{49}=\dfrac{1}{7}$$

In a box,
       $${ B }_{ 1 }=1R,2W$$
       $${ B }_{ 2 }=2R,3W$$
and $${ B }_{ 3 }=3R,4W$$

Let, $${E}_{1}= $$ Student does not know the answer
       $${E}_{2}= $$ Student knows the answer
and $$E=$$ Student answer correctly.

Let $$S$$$$=$$ Event that person is smoker
$$NS$$ $$=$$ Event that person is non-smoker
$$D=$$ Event that death is due to lung cancer.
Now, probability of death due to cancer that a person is a smoker,

The diagrammatic representation of the problem is shown above.