3 Marks Question - Page 4 - MATHS STD 12 Science Questions - Vidyadip

Questions · Page 4 of 4

3 Marks Question

Question 1513 Marks

Two dice are thrown. Find the probability that the numbers appeared has the sum 8, if it is known that the second die always exhibits 4.

Answer

Two dice are thrown.
A = Sum on the dice is 8
A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
B = Second die always exhibits 4
B = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)}
$(\text{A}\cap\text{B})=\{(4, 4)\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{1}{6}$
Required probability $=\frac{1}{6}$

View full question & answer→

Question 1523 Marks

If A and B be two events such that $\text{P(A)}=\frac{1}{4},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{2},$ show that A and B are independent events.

Answer

$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{4}+\frac{1}{3}-\frac{1}{2}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{3+4-6}{12}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{12}=\frac{1}{4}\times\frac{1}{3}=\text{P(A)}\text{ P(B)}$
Thus, A and B are independent events.

View full question & answer→

Question 1533 Marks

If A and B are events such that P(A) = 0.6, P(B) = 0.3 and $\text{P}(\text{A}\cap\text{B})=0.2$ find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ and $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big).$

Answer

Given:
P(A) = 0.6
P(B) = 0.3
$\text{P}(\text{A}\cap\text{B})=0.2$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{0.2}{0.3}=\frac{2}{3}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$\Rightarrow\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{0.2}{0.6}=\frac{1}{3}$

View full question & answer→

Question 1543 Marks

Determine P(E|F) : A coin is tossed three times.
E : at most two tails, F : at least one tail.

Answer

E : at most two tails
$\text{E}=(\text{HTT, THT, TTH, HHT, HTH, THH, HHH})$$\ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \text{n}(\text{E})=7$
$\text{P}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{7}{8}$
F : at least one tail
$\text{F}=(\text{TTT, HTT, THT, TTH, HHT, HTH, THH})\ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \ \text{n}(\text{F})=7$
$\text{P}\left(\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{7}{8}$
$\therefore\ \ \ \ \ \ \ \text{E}\cap\text{F}=\left(\text{HTT, THT, TTH, HHT, HTH, THH}\right)\Rightarrow\ \ \ \text{n}\left(\text{E}\cap\text{F}\right)=6$
$ \therefore\ \ \ \ \ \ \ \text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\cap\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{6}{8}$
$\text{And}\ \ \ \ \text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\cap\text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}$

View full question & answer→

Question 1553 Marks

Find the mean and standard deviation of the following probability distributions:

$x_i$	1	2	3	4
$p_i$	0.4	0.3	0.2	0.1

Answer

$x_i$	$p_i$	$p_ix_i$	$p_ix_i^2$
1	0.3	0.4	0.4
2	0.1	0.6	1.2
3	0.1	0.6	1.8
4	0.3	0.4	1.6
		$\sum\text{p}_\text{i}\text{x}_\text{i}=2$	$\sum\text{p}_\text{i}\text{x}_\text{i}^2=5$

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=2$
Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$
$=5-2^2$
$=1$
Step Deviation $=\sqrt{\text{Variance}}$
$=\sqrt{1}$
$=1$

View full question & answer→

Question 1563 Marks

Five defective bolts are acciedently mixed with twenty good ones. If four bolts are drawn at random from this lot, find the probability distribution of the number of defective bolts.

Answer

Here, 5 defective and 20 non-defective bolts. Let X denote the number of defective bolts drawn out of four bolts drawn. So, X can have values 1, 2, 3, 4.
P(X = 0)
$=\frac{\text{}^{20}\text{C}_4}{\text{}^{25}\text{C}_4}$
$=\frac{4845}{12650}$
$=\frac{969}{2530}$
P(X = 1)
$=\frac{\text{}^{5}\text{C}_1\times\text{}^{20}\text{C}_3}{\text{}^{25}\text{C}_4}$
$=\frac{5700}{12650}$
$=\frac{114}{253}$
P(X = 2)
$=\frac{\text{}^{5}\text{C}_2\times\text{}^{20}\text{C}_2}{\text{}^{25}\text{C}_4}$
$=\frac{1900}{12650}$
$=\frac{38}{253}$
P(X = 1)
$=\frac{\text{}^{5}\text{C}_3\times\text{}^{20}\text{C}_1}{\text{}^{25}\text{C}_4}$
$=\frac{200}{12650}$
$=\frac{4}{253}$
P(X = 4)
$=\frac{\text{}^{5}\text{C}_4}{\text{}^{25}\text{C}_4}$
$=\frac{5}{12650}$
$=\frac{1}{2530}$

View full question & answer→

Question 1573 Marks

A coin is tossed three times. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ in each of the following:
A = At least two heads,
B = At most two heads.

Answer

Sample space for three coins is given by {HHH, HTH, THH, TTH, HHT, HTT, THT, TTT} A = At least two heads A = {HHT, HHT, HTH, THH} B = At most two heads B = {HHT, HTT, THT, TTT, HTH, THH, TTH} $(\text{A}\cap\text{B})=\{\text{HHT, HTH, THH}\}$ $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$ $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{3}{7}$Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{3}{7}$

View full question & answer→

Question 1583 Marks

A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Answer

Let A, B, and C be the respective events that the first, second, and third drawn orange is good.Therefore, probability that first drawn orange is good, $\text{P}(\text{A})=\frac{12}{15}$
The oranges are not replaced.
Therefore, probability of getting second orange good, $\text{P}(\text{B})=\frac{11}{14}$
Similarly, probability of getting third orange good, $\text{P}(\text{C})=\frac{10}{13}$
The box is approved for sale, if all the three oranges are good.
Thus, probability of getting all the oranges good $=\frac{12}{15}\times\frac{11}{14}\times\frac{10}{13}=\frac{44}{91}$
Therefore, the probability that the box is approved for sale is $\frac{44}{91}.$

View full question & answer→

Question 1593 Marks

An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the sum of the numbers obtained is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered $2, 3, 4, ....., 12$ is picked and the number on the card is noted. What is the probability that the noted number is either $7$ or $8$?

Answer

Let $E_1, E_2$ and $A$ be the events as defined below:
$E_1$ = The coin shows a head
$E_2$ = The coin shows a head
$A$ = The noted number is 7 or 8
$\therefore\ \text{P}(\text{E})_1=\frac{1}{2}$
$\text{P}(\text{E})_2=\frac{1}{2}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{11}{36}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{2}{11}$
Using the law of total probability, we get
Required probability $=\text{P(A)}=\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{1}{2}\times\frac{11}{36}+\frac{1}{2}\times\frac{2}{11}$
$=\frac{11}{72}+\frac{1}{11}$
$=\frac{121+72}{792}$
$=\frac{193}{792}$

View full question & answer→

Question 1603 Marks

A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes and, hence, find its mean.

Answer

Let n and p be the parameters of binomial distribution,
Given, $\text{n}=6$
$\text{Mean + Variance}=\frac{10}{3}$
$\text{np + npq}=\frac{10}{3}$
$\text{6p}+\text{6pq}=\frac{10}{3}$
$\text{6p(1 + q})=\frac{10}{3}$
$6(1-\text{q})(1+\text{q})=\frac{10}{3}$ [Since p + q = 1]
$6(1-\text{q}^2)=\frac{10}{3}$
$1-\text{q}^2=\frac{10}{18}$
$-\text{q}^2=\frac{5}{9}-1$
$-\text{q}^2=-\frac{4}{9}$
$\text{q}^2=\frac{4}{9}$
$\text{q}=\frac{2}{3}$
$\text{p}=1-\text{q}$
$=1-\frac{2}{3}$
$\text{p}=\frac{1}{3}$
Hence, the binomial distribution is given by,
$\text{P(X = r})=\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{6-\text{r}}$
as $\text{r}=0,1,2\dots6$

View full question & answer→

Question 1613 Marks

A die is thrown three times. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$, if
A = 4 appears on the third toss,
B = 6 and 5 appear respectively on first two tosses.

Answer

Consider the given events.
A = Getting 4 on third throw
B = Getting 6 on first throw and and 5 on second throw
Clearly,
A = {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4), (2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4), ...... (6, 1, 4), (6, 2 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
Now,
$(\text{A}\cap\text{B})=\{(6, 5, 4)\}$
$\therefore\ \text{Required probability} = \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{1}{6}$
$\therefore\ \text{Required probability} = \text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{A})}=\frac{1}{36}$

View full question & answer→

Question 1623 Marks

A coin is tossed three times, if head occurs on first two tosses, find the probability of getting head on third toss.

Answer

Consider the given events.
A = Getting head on third toss
B = Getting head on first two tosses
Clearly,
A = {(H, H, H), (H, T, H), (T, H, H), (T, T, H)}
B = {(H, H, H), (H, H, T)}
Now,
$\text{A}\cap\text{B}=\{\text{H},\text{H},\text{H}\}$
$\therefore\text{Required probability}=\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{1}{2}$

View full question & answer→

Question 1633 Marks

In eight throws of a die, 5 or 6 is considered a success. Find the mean number of successes and the standard deviation.

Answer

Let p be the probability of success in a singe throw of die
$\text{p}=\frac{2}{6}$ [Since success is occurance of 5 or 6]
$\text{p}=\frac{1}{3}$
$\text{q}=1-\frac{1}{3}$ [Since p + q = 1]
$\text{q}=\frac{2}{3}$
Given, $\text{n}=8$
$\text{Mean = np}$
$=\frac{8}{3}$
$=2.66$
$\text{Standard deviation}=\sqrt{\text{npq}}$
$=\sqrt{8\times\frac{1}{3}\times\frac{2}{3}}$
$=\frac{4}{3}$
$=1.33$
$\text{Mean}=2.66,\text{Standard deviation}=1.33$

View full question & answer→

Question 1643 Marks

Four cards are successively drawn without replacement from a deck of $52$ playing cards. What is the probability that all the four cards are kings?

Answer

Let $E_1, E_2, E_3$ and $E_4$ are the events that the first, second, third and fourth card is king, respectively.
$\therefore\text{P}(\text{E}_1\cap\text{E}_2\cup\text{E}_3\cap\text{E}_4)$
$=\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}_2}{\text{E}_1}\Big)\cdot\text{P}\Big(\frac{\text{E}_3}{\text{E}_1\cap\text{E}_2}\Big)\cdot\text{P}\Big[\frac{\text{E}_4}{(\text{E}_1\cap\text{E}_2\cup\text{E}_3\cap\text{E}_4)}\Big]$
$=\frac{4}{52}\cdot\frac{3}{51}\cdot\frac{2}{50}\cdot\frac{1}{49}$
$=\frac{1}{13\times17\times25\times49}=\frac{1}{270725}$

View full question & answer→

Question 1653 Marks

There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads $75\%$ of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Answer

Let $E_1$= a two headed coin, $E_2$= a biased coin, $E_3$ = an unbiased coin and A = A head is shown$\text{Now}\ \ \text{P}(\text{E}_1)=\frac{1}{3},\ \text{P}(\text{E}_2)=\frac{1}{3},\ \text{P}(\text{E}_3)=\frac{1}{3}$
$\text{P}(\text{A}|\text{E}_1)=1,\ \text{P}(\text{A}|\text{E}_2)=\frac{75}{100}=\frac{3}{4}\ \text{and}\ \text{P}(\text{A}|{\text{E}_3})=\frac{1}{2}$
Therefore, by Bayes’ theorem,
$\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{A}|\text{E}_2)+\text{P}(\text{E}_3)\text{P}(\text{A}|\text{E}_3)}$
$=\frac{\frac{1}{3}\times1}{\frac{1}{3}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{3}\times\frac{1}{2}}=\frac{{4}}{{1}+\frac{3}{4}+\frac{1}{2}}=\frac{4}{4+3+2}=\frac{4}{9}$

View full question & answer→

Question 1663 Marks

There are 3 red and 5 black balls in bag 'A'; and 2 red and 3 black balls in bag 'B'. One ball is drawn from bag 'A' and two from bag 'B'. Find the probability that out of the 3 balls drawn one is red and 2 are black.

Answer

It si givem that bag A contains 3 red and 5 balck balls (3R, 5B) and bag B contains 2 red and 3 black balls (2R, 3B).
Now,
P(One red and 2 black) = P(one red from bag A and two black from bag B) + P(black ball from bag A and remaining balls from bag B)
$=\frac{3}{8}\times\frac{3}{5}\times\frac{2}{4}+\frac{5}{8}\times\frac{2}{5}\times\frac{3}{4}\times2$
$=\frac{9}{80}+\frac{30}{80}$
$=\frac{39}{80}$
Note: 2 is multiplied by second term because there are two ways to select red and black balls from bag B.
While the first way is to pick black ball first, followed by red, the second way is to pick red ball first, followed by black.

View full question & answer→

Question 1673 Marks

Given that the events A and B are such that $\text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{A}\cup\text{B})=\frac{3}{5}$ and P(B) = p. Find p if they are (i) mutually exclusive (ii) independent.

Answer

It is given that $\text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{A}\cup\text{B})=\frac{3}{5},\ \text{and}\ \text{P}(\text{B})=\text{p}$

When A and B are mutually exclusive, $\text{A}\cap\text{B}=\phi$

$\therefore\ \text{P}(\text{A}\cap\text{B})=0$

It is known that, $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$

$\Rightarrow\frac{3}{5}=\frac{1}{2}+\text{p}-0$

$\Rightarrow\ \text{p}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$

When A and B are independent, $\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})\cdot\text{P}(\text{B})=\frac{1}{2}\text{p}$

It is known that, $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$

$\Rightarrow\ \frac{3}{5}=\frac{1}{2}+\text{p}-\frac{1}{2}\text{p}$

$\Rightarrow\frac{3}{5}=\frac{1}{2}+\frac{\text{p}}{2}$

$\Rightarrow\ \frac{\text{p}}{2}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$

$\Rightarrow\text{p}=\frac{2}{10}=\frac{1}{5}$

View full question & answer→

Question 1683 Marks

A die is tossed twice. A 'success' is getting an even number on a toss. Find the variance of number of successes.

Answer

Let p be the probability of getting an even number on the toss when a dice is thrown.
Let q be the probability of not getting an even number on the toss when a dice is thrown.
Then $\text{p}=\frac{3}{2}=\frac{1}{2}$ and $\text{q}=\frac{1}{2}$
Clearly, X follows binomial distribution with $\text{n}=2,\text{p}=\frac{1}{2}.$
$\therefore\text{Variance = npq}=2\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{2}$

View full question & answer→

Question 1693 Marks

The probability is 0.02 that an item produced by a factory is defective. A shipment of 10,000 items is sent to its warehouse. Find the expected number of defective items and the standard deviation.

Answer

Let p denote the probability of a defective item produced in the factory, so
$\text{p}=0.02$
$=\frac{2}{100}$
$\text{p}=\frac{1}{50}$
$\text{q}=1-\frac{1}{50}$ [Since p + q = 1]
$=\frac{49}{50}$
Given $\text{n}=10,000$
Expected number of defective item $=\text{np}$
$=10000\times\frac{1}{50}$
$=200$
$\text{Standard deviation}=\sqrt{\text{npq}}$
$=\sqrt{10000\times\frac{1}{50}\times\frac{49}{50}}$
$=14$
$\text{Expected No. of defective items}=200$
$\text{Standard deviation}=14$

View full question & answer→

Question 1703 Marks

Three machines $\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3$ in a certain factory produce $50 \%, 25 \%$ and $25 \%$, respectively, of the total daily output of electric bulbs. It is known that $4 \%$ of the tubes produced one each of the machines $\mathrm{E}_1$ and $\mathrm{E}_2$ are defective, and that $5 \%$ of those produced on $E_3$ are defective. If one tube is picked up at random from a day's production, then calculate the probability that it is defective.

Answer

Let D be the event that the picked up tube is defective.
Let $A_1, A_2$ and $A_3$ be the events that the tube is produced on macjines $E_1, E_2$ and $E_3$ respectively.
$P(D) = P(A_1) P(D|A_1) + P(A_2) P(D|A_2) + P(A_3) P(D|A_3)$ .....(i)
$\text{P}(\text{A}_1)=\frac{50}{100}=\frac{1}{2},\text{P}(\text{A}_2)=\frac{25}{100}=\frac{1}{4},\text{P}(\text{A}_3)=\frac{25}{100}=\frac{1}{4}$
$\text{P}(\text{D}|\text{A}_1)=\text{P}(\text{D}|\text{A}_2)=\frac{4}{100}=\frac{1}{25}$
$\text{P}(\text{D}|\text{A}_3)=\frac{5}{100}=\frac{1}{20}$
Putting these values in (i), we get
$\text{P(D)}=\frac{1}{2}\times\frac{1}{25}+\frac{1}{4}\times\frac{1}{25}+\frac{1}{4}\times\frac{1}{20}$
$\text{P(D)}=\frac{17}{400}$

View full question & answer→

Question 1713 Marks

Let $X$ be a random variable which assumes values $x_1, x_2, x_3, x_4$ such that $2 P\left(X=x_1\right)=3 P\left(X=x_2\right)=P\left(X=x_3\right)=5 P(X=$ $x_4$ ). Find the probability distribution of $X$.

Answer

Here,$2P(x_1) = 3P(X_2) = P(x_3) = 5P(x_4)$
Let $P(x_3) = a$
$2\text{P}(\text{x}_1)=\text{P}(\text{x}_3)\Rightarrow\text{P}(\text{x}_1)=\frac{\text{a}}{2}$
$3\text{P}(\text{x}_2)=\text{P}(\text{x}_3)\Rightarrow\text{P}(\text{x}_2)=\frac{\text{a}}{3}$
$5\text{P}(\text{x}_4)=\text{P}(\text{x}_3)\Rightarrow\text{P}(\text{x}_4)=\frac{\text{a}}{5}$
Since, $\text{P}(\text{x}_1)+\text{P}(\text{x}_2)+\text{P}(\text{x}_3)+\text{P}(\text{x}_4)=1$
$\Rightarrow\frac{\text{a}}{2}+\frac{\text{a}}{3}+\frac{\text{a}}{1}+\frac{\text{a}}{5}=1$
$\Rightarrow\frac{15\text{a}+10\text{a}+30\text{a}+6\text{a}}{30}=1$
$\Rightarrow61\text{a}=30$
$\Rightarrow\text{a}=\frac{30}{61}$
So,

$\text{X}:$	$\text{x}_1$	$\text{x}_2$	$\text{x}_3$	$\text{x}_4$
$\text{P}(\text{X}):$	$\frac{15}{61}$	$\frac{10}{61}$	$\frac{30}{61}$	$\frac{6}{61}$

View full question & answer→

Question 1723 Marks

For a loaded die, the probabilities of outcome are given as under:
P(1) = P(2) = 0.2, P(3) = P(5) = P(6) = 0.1 and P(4) = 0.3. the die is thrown two times. If the die were fair, determine whether or not the events A and B are independent.

Answer

We have $A = A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}$
$\therefore$ n(A) = 6 and $n(S) = 6^2 = 36$
$\therefore\text{P}(\text{A})=\frac{\text{n}(\text{A})}{\text{n}(\text{S})}$
$=\frac{6}{36}=\frac{1}{6}$
and $B = {(4, 6), (6, 4), (5, 5), (6, 5), (5, 6), (6, 6)}$
$\Rightarrow n(B) = 6$
$\therefore\text{P}(\text{B})=\frac{\text{n}(\text{B})}{\text{n}(\text{S})}$
$=\frac{6}{36}=\frac{1}{6}$
Also, $\text{A}\cap\text{B}=\left\{(5,5),(6,6)\right\}$
$\therefore\text{P}(\text{A}\cap\text{B})=\frac{2}{36}=\frac{1}{8}$
Also, $\text{P}(\text{A})\cdot\text{P}(\text{B})=\frac{1}{36}$
Thus, $\text{P}(\text{A}\cap\text{B})\neq\text{P}(\text{A})\cdot\text{P}(\text{B})$
So, A and B are not independent events.

View full question & answer→

Question 1733 Marks

The items produced by a company contain 10% defective items. Show that the probability of getting 2 defective items in a sample of 8 items is $\frac{28\times9^6}{10^8}$.

Answer

Let X denote the number of defective items in the items produced by the company.
Then, X follows binomial distribution with n = 8.
$\text{p}=10\%=\frac{1}{10}$
$\text{q}=1-\text{p}=\frac{9}{10}$
Hence, the distribution is given by
$\text{P(X = r)}=\text{ }^8\text{C}_{\text{r}}\big(\frac{1}{10}\big)^{\text{r}}\big(\frac{9}{10}\big)^{8-\text{r}}$
Prob of getting 2 defective items $=\text{P}(\text{X}=2)$
$=\text{ }^8\text{C}_2\big(\frac{1}{10}\big)^2\big(\frac{9}{10}\big)^{8-2}$
$=\frac{28\text{ x }9^6}{10^8}$

View full question & answer→

Question 1743 Marks

The mean of a binomial distribution is 20 and the standard deviation 4. Calculate the parameters of the binomial distribution.

Answer

Given that mean, i.e. $\text{np}=20\dots(1)$
and standard deviation, i.e. $\text{npq}=4$
$\sqrt{\text{npq}}=4$
$\Rightarrow\text{npq}=16\dots(2)$
Dividing eq. (2) by eq. (1), we get
$\text{q}=\frac{16}{20}=\frac{4}{5}$
and $\text{p}=\frac{1}{5};$
$\therefore\text{n}=\frac{\text{Mean}}{\text{p}}=100$
$\text{P(X = r})=\text{ }^{100}\text{C}_{\text{r}}\big(\frac{1}5{})^{\text{r}}\big(\frac{4}{5}\big)^{100-\text{r}},\text{r}=0,1,2\dots100$
Therefore, the parameters are $\text{n}=100$ and $\text{p}=\frac{1}{5}$

View full question & answer→

Question 1753 Marks

A box has $5$ blue and $4$ red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of second ball being blue?

Answer

A box $5$ blue and $4$ red balls.
Let $E_1$ is the event that first ball drawn is blue, $E_2$ is the event that first ball.
drawn is red and E is the event that second ball drawn is blue.
$\therefore\text{P}(\text{E})=\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)$
$=\frac{5}{9}\cdot\frac{4}{8}+\frac{4}{9}\cdot\frac{5}{8}$
$=\frac{20}{72}+\frac{20}{72}=\frac{40}{72}$
$=\frac{5}{9}$

View full question & answer→

Question 1763 Marks

A bag contains 5 white, 7 red and 3 black balls. If three balls are drawn one by one without replacement, find the probability that none is red.

Answer

Bag contains 5 white, 7 red and 3 black balls.
Total number of balls = 15
Three balls are drawn without replacement
A = first ball is red
B = Second ball is red
C = Third balls is red
P (Three balls are drawn, non is red)
$=\text{P}(\overline{\text{A}})\text{P}\Big(\overline{\frac{\text{B}}{\text{A}}}\Big)\text{P}\Big(\frac{\overline{\text{C}}}{\text{A}\cap\text{B}}\Big)$
$=\frac{8}{15}\times\frac{7}{14}\times\frac{6}{13}$
[Since, number of non red balls = 5 + 3 = 8]
$=\frac{8}{65}$
Required probability $=\frac{8}{65}$

View full question & answer→

Question 1773 Marks

Ten eggs are drawn successively, with replacement, from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.

Answer

Let X be the number of defective eggs drawn from 10 eggs.
Then, X follows a binomial distribution with n = 10
Let p be the probability that a drawn egg is defective.
$\therefore\text{p}=10\%=\frac{1}{10},\text{q}=\frac{9}{10}$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^{10}\text{C}_{\text{r}}\big(\frac{1}{10}\big)^{\text{r}}\big(\frac{9}{10}\big)^{10-\text{r}},\text{r}=0,1,2\dots10$
$\text{P(there is at least one defectiv egg})=\text{P(X}\geq1)$
$=1-\text{P(X}=0)$
$=1-\text{ }^{10}\text{C}_0\big(\frac{1}{10}\big)^0\big(\frac{9}{10}\big)^{10-0}$
$=1-\big(\frac{9}{10}\big)^{10}$
$=1-\frac{9^{10}}{10^{10}}$

View full question & answer→

Question 1783 Marks

Find the binomial distribution when the sum of its mean and variance for 5 trials is 4.8.

Answer

Number of trials in the binomial distribution = 5
If p is the probability for success, then
$\text{np + npq}=4.8$
Or $\Rightarrow5\text{p}+5\text{p}(1-\text{p})=4.8$
By factorising, we get
$\big(\text{p}-0.8\big)\big(\text{p}-1.2)=0$
As p cannot exceed 1,
$\text{p}=0.8$ or $\frac{4}{5}$
and $\text{q}=1-\text{p}=\frac{1}{5}$
$\therefore\text{P(X = r})=\text{ }^{5}\text{C}_{\text{r}}\big(\frac{4}{5}\big)^{\text{r}}\big(\frac{1}{5}\big)^{5-\text{r}},\text{r}=0,1,2,\dots5$

View full question & answer→

Question 1793 Marks

Find the mean and standard deviation of the following probability distributions:

$x_i$	0	1	3	5
$p_i$	0.2	0.5	0.2	0.1

Answer

$x_i$	$p_i$	$p_ix_i$	$p_ix_i^2$
0	0.2	0	0
1	0.5	0.5	0.5
3	0.2	0.6	1.8
5	0.1	0.5	2.5
		$\sum\text{p}_\text{i}\text{x}_\text{i}=1.6$	$\sum\text{p}_\text{i}\text{x}_\text{i}^2=4.8$

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=1.6$
Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$
$=4.8-1.6^2$
$=2.24$
Step Deviation $=\sqrt{\text{Variance}}$
$=\sqrt{2.24}$
$=1.497$

View full question & answer→

Question 1803 Marks

Three cards are drawn with replacement from a well shuffled pack of 52 cards. Find the probability that the cards are a king, a queen and a jack.

Answer

Three cards are drawn with replacement fron a pack of cards
There are 4 Kings, 4 Queens, 5 Jacks.
P (1 King, 1 Queen, 1 Jack)
$=\text{P}\big[(\text{K}\cap\text{Q}\cap\text{J})\cup(\text{K}\cap\text{J}\cap\text{Q})\cup(\text{J}\cap\text{K}\cap\text{Q})\\\cup(\text{J}\cap\text{Q}\cap\text{K})\cup(\text{Q}\cap\text{K}\cap\text{L})\cup(\text{Q}\cap\text{J}\cap\text{K})\big]$
$=\text{P}(\text{K}\cap\text{Q}\cap\text{J})+\text{P}(\text{K}\cap\text{J}\cap\text{Q})+\text{P}(\text{J}\cap\text{K}\cap\text{Q})\\+\text{P}(\text{J}\cap\text{Q}\cap\text{K})+\text{P}(\text{Q}\cap\text{K}\cap\text{L})+\text{P}(\text{Q}\cap\text{J}\cap\text{K})$
$=\text{P}(\text{K}) \text{P}(\text{Q}) \text{P}(\text{J})+\text{P}(\text{K}) \text{P}(\text{J}) \text{P}(\text{Q})+\text{P}(\text{J}) \text{P}(\text{K}) \text{P}(\text{Q}) \\ +\text{P}(\text{J}) \text{P}(\text{Q}) \text{P}(\text{K})+\text{P}(\text{Q}) \text{P}(\text{K}) \text{P}(\text{L})+\text{P}(\text{Q}) \text{P}(\text{J}) (\text{K})$
$=\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52} \\ +\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}$
$=\frac{6}{13\times13\times13}$
$=\frac{6}{2197}$
Required probability $=\frac{6}{2197}$

View full question & answer→

Question 1813 Marks

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$ ).

Answer

There are four entries in determinant of 2 x 2 order. Each entry may be filled up in two ways with 0 or 1.
$\therefore$ number of determinants that can be formed = $2^4 = 16$
$\therefore$ total number of cases = 16
The value of determinant is positive in the cases
$ \begin{vmatrix} 1\ 0\\0\ 1 \end{vmatrix}, \begin{vmatrix} 1\ 0\\1\ 1 \end{vmatrix}, \begin{vmatrix} 1\ 1\\0\ 1 \end{vmatrix},$
$\therefore$ number of favourable cases = 3
$\therefore$ the probability that the determinant is positive $=\frac{3}{16}.$

View full question & answer→

Question 1823 Marks

A fair die is tossed. Let X denote twise the number appearing. Find the probability distribution, mean and variance of X.

Answer

$\text{x}_\text{i}$	$\text{p}_\text{i}$	$\text{p}_\text{i}\text{x}_\text{i}$	$\text{p}_\text{i}\text{x}_\text{i}^2$
$2$	$\frac{1}{6}$	$\frac{2}{6}$	$\frac{4}{6}$
$4$	$\frac{1}{6}$	$\frac{4}{6}$	$\frac{16}{6}$
$6$	$\frac{1}{6}$	$\frac{6}{6}$	$\frac{36}{6}$
$8$	$\frac{1}{6}$	$\frac{8}{6}$	$\frac{64}{6}$
$10$	$\frac{1}{6}$	$\frac{10}{6}$	$\frac{100}{6}$
$12$	$\frac{1}{6}$	$\frac{12}{6}$	$\frac{144}{6}$
		$\sum\text{x}_\text{i}\text{p}_\text{i}=7$	$\sum\text{x}_\text{i}\text{p}_\text{i}^2=\frac{364}{6}$

Mean $=\sum\text{x}_\text{i}\text{p}_\text{i}=7$
Variance $=\sum\text{x}_\text{i}\text{p}_\text{i}^2-(\text{Mean})^2$
$=60.7-49$
$=11.7$

View full question & answer→

Question 1833 Marks

Find the mean and standard deviation of the following probability distributions:

$x_i$	1	2	3	4
$p_i$	0.4	0.1	0.2	0.3

Answer

$x_i$	$p_i$	$x_ip_i$	$x_i^2p_i$
1	0.4	0.4	0.4
3	0.1	0.3	0.9
4	0.2	0.8	3.2
5	0.3	1.5	7.5
		$\sum\text{xp}=3$	$\sum\text{x}^2\text{p}=12$

Mean $=\sum\text{xp}=3$
Standard deviation $=\sqrt{\sum\text{x}^2\text{p}-(\text{mean})^2}$
$=\sqrt{12-(3)^2}$
$=\sqrt3$
Standard deviation = 1.732

View full question & answer→

Generate Paper Free All Probability topics

3 Marks Question - Page 4 - MATHS STD 12 Science Questions - Vidyadip