Probability Test - 25

We define the following events:
$${ A }_{ 1 }:$$ Selecting a pair of consecutive letters from the word $$LONDON$$
$${ A }_{ 2 }:$$ Selecting a pair of consecutive letters from the word $$CLIFTON$$
$$E:$$ Selecting a pair of letters $$ON$$
Then $$\displaystyle P\left( { A }_{ 1 }\cap E \right) =\frac { 2 }{ 5 } $$, as there are $$5$$ pairs of consecutive letters out of which $$2$$ are $$ON$$.
$$\displaystyle P\left( { A }_{ 2 }\cap E \right) =\frac { 1 }{ 6 } $$, as there are $$6$$ pairs of consecutive letters of which $$1$$ is $$ON$$.
So, required probability $$\displaystyle =P\left( \frac { { A }_{ 1 } }{ E }  \right) $$
$$\displaystyle \Rightarrow P\left( \frac { { A }_{ 1 } }{ E }  \right) =\frac { P\left( { A }_{ 1 }\cap E \right)  }{ P\left( { A }_{ 1 }\cap E \right) +P\left( { A }_{ 1 }\cap E \right)  } =\dfrac { \dfrac { 2 }{ 5 }  }{ \dfrac { 2 }{ 5 } +\dfrac { 1 }{ 6 }  } =\dfrac { 12 }{ 17 } $$

$$\mathbf{{\text{Step -1: Stating all the possibilities}}{\text{.}}}$$

                 $${\text{Writing down all the possibilities that the target is hit by the second plane}}{\text{.}}$$

                 $${\text{First plane fails and then the second one succeed}}{\text{.}}$$

                 $${\text{The first one fails, then second fails, the again first one fails, and then the second one succeed}}{\text{.}}$$

                 $$\text{Similarly, we can observe that all cases in which the second plane succeed }$$ 

                 $${\text{at last will be considered}}{\text{.}}$$

$$\mathbf{{\text{Step -2: Finding the probability}}{\text{.}}}$$

                 $${\text{We know that the probability of first plane hitting correctly is 0}}{\text{.3}}$$

                 $${\text{and that of the second plane is 0}}{\text{.2}}{\text{.}}$$

                 $${\text{So let A be the event when first plane succeeds and B be the event when second plane succeeds}}{\text{.}}$$

                 $${\text{P(A) = 0}}{\text{.3 and P(B) = 0}}{\text{.2}}$$

                 $${\text{Let E be the event that the target is hit by the second plane}}{\text{.So we can write,}}$$

                 $${\text{P(E) = P(1 - A)P(B) + P(1 - A)P(1 - B)P(1 - A)P(B) + }}$$ 

                                $${\text{P(1 - A)P(1 - B)P(1 - A)P(1 - A)P(1 - B)P(B) + }}......\infty $$

                 $${\text{The expression written above represent that first fails and second succeeded,  }}$$

                 $${\text{first fails, second, fails, again first fails and then second succeeded,}}....$$

                 $${\text{So P(E) = 0}}{\text{.7}} \times 0.2 + 0.7 \times 0.8 \times 0.7 \times 0.2 + 0.7 \times 0.8 \times 0.7 \times 0.8 \times 0.7 \times 0.2 + .....\infty $$

                 $${\text{P(E) = (0}}{\text{.7}} \times 0.2)(1 + 0.8 \times 0.7 + 0.8 \times 0.7 \times 0.8 \times 0.7 + .....\infty )$$

                 $${\text{From this we can observe that }}1 + 0.8 \times 0.7 + 0.8 \times 0.7 \times 0.8 \times 0.7 + .....\infty {\text{ forms a GP with }}$$

                 $${\text{infinite terms and common ratio(0}}{\text{.56) < 1}}{\text{.}}$$

                 $${\text{As we know that sum of such a GP = }}\dfrac{{\text{a}}}{{1 - {\text{r}}}}$$

                 $${\text{P(E) = (0}}{\text{.14)}} \times \dfrac{{\text{1}}}{{1 - 0.56}}$$

                 $${\text{P(E) = }}\dfrac{{0.14}}{{0.44}}$$$${\text{ = }}\dfrac{7}{{22}}.$$

$$\mathbf{{\text{}}{\text{ Thus, the probability that the target is hit by the second plane is }}\dfrac{7}{{22}}{\text{ = 0}}{\text{.32}}{\text{.}}}$$

Since $$\displaystyle P\left ( \frac{E_{1}}{E_{2}} \right )=P\left ( E_{1} \right )$$
$$\Rightarrow $$ $$E_{1}$$ and $$E_{2}$$ are independent of each other
Also since $$P(E_{1}\cup E_{2})=P\left ( E_{1} \right )+P\left ( E_{2} \right )-P\left ( E_{1} \right )|P\left ( E_{2} \right )\neq 1$$
Hence events are not exhaustive. Independent events can't be mutually exclusive.
Hence only (i) is correct
Further since $$E_{1}$$ & $$E_{2}$$ are independent; $$E_{1}$$ and $$\bar{E_{2}}$$ or $$\bar{E_{1}},$$ $$E_{2}$$ are $$\bar{E_{1}}, \bar{E_{2}}$$ are also independent.
Hence $$\displaystyle P\left ( \frac{\bar{E_{1}}}{E_{2}} \right )=P\left ( \bar{E_{1}} \right )=\frac{3}{4}$$ and $$\displaystyle P\left ( \frac{E_{2}}{\bar{E_{1}}} \right )=P\left ( E_{2} \right )=\frac{1}{2}$$

Let $$b$$ and $$g$$ represent the boy and the girl child respectively. If a family has two children, the sample space will be

Out of the $$ 70 $$ numbers, numbers that are a multiple of $$ 5 $$ or $$ 7 $$ are $$ 5, 7, 10, 14, 15, 20, 21, 25, 28, 30, 35, 40, 42, 45, 49, 50, 55, 56, 60, 63, 65, 70 $$


So, probability that the number is even $$ = \dfrac {22}{70} = \dfrac {22}{70} = \dfrac {11}{35} $$

If two coins are tossed once, then the sample space $$S$$ is

Total number of vowels in English alphabet $$ = 5 $$ which are $$ a,e,i,o,u $$


So, probability of $$ u $$ when a vowel is selected $$ = \dfrac {Number \ of   \ favourable \  outcomes}{Total \ number \   of \ outcomes} = \dfrac {1}{5} $$

A coin is tossed six times and $$X$$ represents the difference between the number of heads and the number of tails.
$$\therefore X\left( 6H,0T \right) =\left| 6-0 \right| =6$$
$$X\left( 5H,1T \right) =\left| 5-1 \right| =4$$
$$X\left( 4H,2T \right) =\left| 4-2 \right| =2$$
$$X\left( 3H,3T \right) =\left| 3-3 \right| =0$$
$$X\left( 2H,4T \right) =\left| 2-4 \right| =2$$
$$X\left( 1H,5T \right) =\left| 1-5 \right| =4$$
$$X\left( 0H,6T \right) =\left| 0-6 \right| =6$$
Thus, the possible values of $$X$$ are $$0, 2, 4$$ and $$6$$.