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MATHS 3 Marks Question

Probability · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 183 questions.

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Question 13 Marks
Determine the binomial distribution for which the mean is $20$ and variance $16.$
Answer
$n . p = 20,$
$n p q = 16$
$\Rightarrow\text{q}=\frac{16}{20}=\frac{4}{5}\text{ }\text{ }\therefore\text{ }\text{ }\text{p}=\frac{1}{5}\text{and n}=100$
OR  The Distribution is $ P(r) = 100_{\text{C}_\text{r}}\Bigg[\frac{4}{5}\Bigg]^\text{100 - r}\cdot\Bigg(\frac{1}{5}\Bigg)^\text{r},\text{ r = 0,1,2,.....,100.}$
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Question 23 Marks
An urn contains 4 red and 7 blue balls. Two balls are drawn at random with replacement. Find the probability of getting:
  1. 2 red balls.
  2. 2 blue balls.
  3. One red and one blue ball.
Answer
  1. P (2 red balls) =$\frac{4}{11}\cdot\frac{4}{11}=\frac{16}{121}$.
  2. P (2 blue balls) = $\frac{7}{11}\cdot\frac{7}{11}=\frac{49}{121}$.
  3. P (one red and one blue ball) = $\frac{4}{11}\cdot\frac{7}{11}\cdot2=\frac{56}{121}$.
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Question 33 Marks
Two dice are thrown together. What is the probability that the sum of the numbers on the two dice is neither 9 nor 11?
Answer
n (s) = 36P (getting the sum of 9) = P{(3,6), (6, 3), (4, 5), (5, 4)} = $\frac{4}{36}=\frac{1}{9}$
P (getting the sum of 11) = P {(5, 6), (6, 5)} = $\frac{2}{36}=\frac{1}{18}$
P (neither 9 or 11) = 1 – P (either 9 or 11) = 1 - $\Big[\frac{1}{9}+\frac{1}{18}\Big]$
= 1 - $\frac{3}{18}=1-\frac{1}{6}=\frac{5}{6}$.
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Question 43 Marks
There are two bags $I$ and $II.$ Bag I contains $3$ white and $4$ red balls and Bag $II$ contains $5$ white and $6$ red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag $II.$
Answer
Let $E_1:$ drawing a ball from bag $I,$
$\therefore\text{P(E}_{1})=\frac{1}{2}$
$E_2:$ drawing a ball from bag $II, $
$\therefore\text{P(E}_{2})=\frac{1}{2}$
$A:$ getting a red ball,
$\therefore\text{P(A/E}_{1})=\frac{4}{7},\text{ P(A/E}_{2})=\frac{6}{11}$
$\therefore$ Required probability $=\  \therefore\text{ P(E}_{2}/\text{A})=\frac{\text{P(E}_{2})\cdot\text{P(A/E}_{2})}{\text{P(E}_{1})\cdot\text{P(A/E}_{1})+\text{P(E}_{2})\cdot\text{P(A/E}_{2})}$
$\frac{\frac{1}{2}\cdot\frac{6}{11}}{\frac{1}{2}\cdot\frac{4}{7}+\frac{1}{2}\cdot\frac{6}{11}}=\frac{21}{43}$.
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Question 53 Marks
Find the mean $\mu $ variance $\sigma ^2$ for the following probability distribution:
$X$ $0$ $1$ $2$ $3$
$P(X)$ $\frac{1}{6}$ $\frac{1}{2}$ $\frac{3}{10}$ $\frac{1}{30}$
Answer
$X:$ $0$ $1$ $2$ $3$ Sum
$P(X):$ $\frac{1}{6}$ $\frac{1}{2}$ $\frac{3}{10}$ $\frac{1}{30}$ $1$
$X^{_{. }}P(X):$ $0$ $\frac{1}{2}$ $\frac{3}{5}$ $\frac{1}{10}$ $\frac{6}{5}$
$X^{2 }P(X):$ $0$ $\frac{1}{2}$ $\frac{6}{5}$ $\frac{1}{10}$ $2$
Mean ($\mu) = \Sigma X P(X)= \frac{6}{5}$ or $1.2$
Variance $(\sigma)=\Sigma X^{2_{. }}P(X)-[\Sigma X ^{_{. }}P(X)]^2= \frac{14}{25}$ OR $0.56.$
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Question 63 Marks
A and B toss a coin alternately till one of them gets a head and wins the game. If A starts first, find the probability that B will win the game.
Answer
P (Head) = $\frac{1}{2}$ P (Tail) = $\frac{1}{2}$ 
P (B wins) = $\text{P}[(\overline{\text{A}}\text{B})\text{ OR }(\overline{\text{A}}{\text{ }}\overline{\text{B}}\text{ } \overline{\text{A}}{\text{ }}\text{B})\text{ OR }(\overline{\text{A}}{\text{ }}\overline{\text{B}}\text{ } \overline{\text{A}}{\text{ }}\overline{\text{B}}\text{ } \overline{\text{A}}{\text{ }}\text{B})\text{ OR }.......]$
$=\Bigg(\frac{1}{2}\cdot\frac{1}{2}\Bigg)+\Bigg(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\Bigg)+\Bigg(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\Bigg)+..........$
$=\frac{1}{4}\cdot\Bigg(1+\frac{1}{4}+\Bigg(\frac{1}{4}\Bigg)^{2}+.......\Bigg)=\frac{1}{4}\Bigg(\frac{1}{1-\frac{1}{4}}\Bigg)=\frac{1}{4}\cdot\frac{4}{3}=\frac{1}{3}.$
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Question 73 Marks
Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of number of jacks.
Answer
$\text{p}=\frac{4}{52}=\frac{1}{13},\text{ q}=\frac{12}{13}$
Number of Jacks (X): 0 1. 2.
P(X): $\Bigg(\frac{12}{13}\Bigg)^{2}$ $2.\frac{1}{13}\cdot\frac{12}{13}$ $\Bigg(\frac{1}{13}\Bigg)^{2}$
  $\frac{144}{169}$ $\frac{24}{169}$ $\frac{1}{169}$
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Question 83 Marks
Find the binomial distribution for which the mean is 4 and variance 3.
Answer
$\text{np} = \text{4 and npq} = 3 $$\therefore \text{q} = \frac{3}{4} \Rightarrow \text{p} = \frac{1}{4} $
$\text{np} = 4 \Rightarrow \text{n} \times\frac{1}{4}= 4 \Rightarrow \text{n} = 16$
$\text{For writing n} = 16, \text{p} = \frac{1}{4}, \text{q} = \frac{3}{4} \text{or P (r)} = 16_\stackrel{{C}}{{r}}\bigg(\frac{1}{4}\bigg)^{16 - r} , r = 0, 1, 2, \dots\dots16$
 
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Question 93 Marks
Find mean $u,$variance $\sigma^{2}$for the following probability distribution:
X 0 1 2 3
P(X) $\frac{1}{8}$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{1}{8}$
 
Answer
$\text{X}{_i}$ 0 1 2 3 Total
$\text{P (X}_{i})$ $\frac{1}{8}$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{1}{8}$ 1
$\text{X}_{i} \text{P X}_{i}$ 0 $\frac{3}{8}$ $\frac{6}{8}$ $\frac{3}{8}$ $\frac{3}{2}$
$\text{Xi}^{2} \text{P (X}_{i})$ 0 $\frac{3}{8}$ $\frac{3}{2}$ $\frac{3}{9}$ $\frac{24}{8} = 3$
$\text{Mean (u)} \Sigma \text{X}_{i} \text{P X}{i} = \frac{3}{2} \text{or 1.5} $
$\text{Variance} (\sigma^{2}) = \Sigma X^{2}_{i} P(X_{i}) - [\Sigma X _{i} P(X_{i}) ]^{2} $
$= 3 - \bigg(\frac{3}{2}\bigg)^{2}$
$3 - \frac{9}{4} = \frac{3}{4} \text{or 0.75}$
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Question 103 Marks
There are two bags I and II. Bag I contains 2 white and 3 red balls and Bag II contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag II.
Answer
Let $\text{E}_{1} :$ drawing ball from Bag I $\Rightarrow P(E_{1}) = \frac{1}{2}$$\text{E}_{2} :$ drawing a ball from Bag II $\Rightarrow P E_{2} = \frac{1}{2}$
$\text{A} :$ Getting a red ball $\Rightarrow P \bigg(\frac{A}{E_1}\bigg) = \frac{3}{5}, P\bigg( \frac{A}{E_{2}} \bigg) = \frac{5}{9}$
$P \bigg( \frac{E_{2}}{A}\bigg) =\frac{P(E_{2}).\bigg(\frac{A}{E_2}\bigg)}{P (E_{1}). P\bigg(\frac{A}{E_{1}}\bigg)+ P \bigg( P(E_{2}) . P\bigg(\frac{A}{E_{2}}\bigg)}$
$= \frac{\frac{1}{2}.\frac{5}{9}}{{\frac{1}{2}}.\frac{3}{5} + \frac{1}{2} .\frac{5}{9}} = \frac{25}{52}$
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Question 113 Marks
A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that it is neither an ace nor a king.
Answer
$\text{P (an Acc)} =\frac{4}{52} = \frac{1}{13}$$\text{P (a king)} =\frac{4}{52} = \frac{1}{13}$
$\therefore \text{P(a king or an Ace)} = \frac{1}{13} + \frac{1}{13} = \frac{2}{13}$
$\therefore \text{P(Neither a king nor an Ace)} = 1 -\frac{2}{13} = \frac{11}{13}$
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Question 123 Marks
Two dice are rolled once. Find the probability that:
  1. the numbers on two dice are different.
  2. the total of numbers on the two dice is at least.
Answer
P(Numbers on two dice are different) $= \frac{30}{36} =\frac {5}{6}$ P (Total of numbers on two dice is atleast 4) $=\frac{33}{36} = \frac{11}{12}$
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Question 133 Marks
A pair of dice is tossed twice. If the random variable X is defined as the number of doublets, find the probability distribution of X.
Answer
Let x denote the number of doublets. Possible doublets are: -(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Clearly X can take values 0, 1, 2
Probability of:-
  1. Getting a doublet $= \frac{1}{6}$
  2. Not getting a doublet $= 1 -\frac{1}{6} = \frac{5}{6}$
P (X = 0) = P (not doublet) $= \frac{5}{6}\times \frac{5}{6} = \frac{25}{36}$
P(X = 1) = P (one doublet and one not doubled) $= 2 \frac{1}{6} \frac{5}{6} = \frac{10}{36}$
P(X = 2) = P (both doublets) $= \frac{1}{6} \frac{1}{6} = \frac{1}{36}$
$\therefore$ Required probability distribution is:-
X
0
1
2
P (X)
$\frac{25}{36}$
$\frac{10}{36}$
$\frac{1}{36}$
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Question 143 Marks
In a factory, which manufactures nuts, machines A, B and C manufacture respectively 25%, 35% and 40% of nuts. Of their outputs, 5, 4 and 2 per cent respectively are defective nuts. A nut is drawn at random from the product and is found to be defective. Find the probability that it is manufactured by machine B.
Answer
$\text{Let the events}A_{1}, A_{2}, A_{3} \text{E be as follows} :$$A_{1} :$ the nut is manufactured by machine A
$A_{2} :$ the nut is manufactured by machine B
$A_{3} :$ the nut is manufactured by machine C
${E:}$ the nut manufactured is defective
$\therefore P(A_{1}) = 0.25, P(A_{2}) = 0.35, P(A_{3}) = 0.40 $
$P (E/A_{1}) = 0.05, P(E/A_{2}) = 0.04, P(E/A_{3}) = 0.02$
$P(A_{2}/E) = \frac{P(A_{2}).P(E/A_{2})}{\sum\limits^{3}_{i-1}P(A_{i}).P(E/A_i)}$
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Question 153 Marks
8% of people in a group are left handed. What is the probability that 2 or more of a random sample of 25 from the group are left handed? $e^{-2} = 0.135$ 
Answer
Here ${\lambda} = np = 25 \times \frac{8}{100} =2 $$\text{Required probability} = P (X\geq{-} 2)$
$= 1 -P (X = 0) - P (X = 1)$
$= 1-\frac{e^{-2}}{1!} - \frac{2e^{-2}}{1!}$
$= 1 - 3e^{-2} = 1 - 3(.135) = 0.595$
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Question 163 Marks
If the mean and variance of a binomial distribution are respectively 9 and 6, find the distribution.
Answer
$\text{Here Mean = np = 9 and variance = npq = 6}$$\therefore q = \frac{6}{9} = \frac{2}{3}$
$\therefore p = 1- q= \frac{1}{3}$
$\text{Again np} = 9$
$\Rightarrow n \times \frac{1}{3} = 9 \Rightarrow n = 27$
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Question 173 Marks
Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?
Answer
Let $\text{E}_1,\ \text{E}_2$ E be the events$\text{E}_1$ : 'A student is residing in hostel'
$\text{E}_2$: 'A student is day scholar',
E : 'A student gets A grade',
$\text{Now},\ \ \ \ \ \text{P}(\text{E}_1)=\frac{60}{100}=\frac{3}{5}\ \text{and}\ \text{P}(\text{E}_2)=\frac{40}{100}=\frac{2}{5}$
$\therefore\ \text{P}(\text{E}|\text{E}_1)=\frac{30}{100}=\frac{3}{10}\ \text{and}\ \text{P}(\text{E}|\text{E}_2)=\frac{20}{100}=\frac{2}{10}$
Required probability $=\text{P}(\text{E}_1|\text{E})$
$=\frac{\text{P}(\text{E}_1)\text{P}(\text{E}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{E}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{E}|\text{E}_2)}$ (By Baye's Theorem)
$=\frac{\frac{3}{5}\times\frac{3}{10}}{\frac{3}{5}\times\frac{3}{10}+\frac{2}{5}\times\frac{2}{10}}=\frac{9}{9+4}=\frac{9}{13}.$
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Question 183 Marks
If the probability of a defective bolt is 0.1, find the (1) mean and (2) standard deviation for the distribution of bolts in a total of 400 bolts.
Answer
Let p denotes the probability of selecting a defective bolt, so
$\text{p}=0.1$
$\text{p}=\frac{1}{10}$
$\text{q}=1-\frac{1}{10}$ [Since p + q = 1]
$\text{q}=-\frac{9}{10}$
Given, $\text{n}=400$
(1)
$\text{Mean = np}$
$=400\times\frac{1}{10}$
$\text{Mean = 40}$
(2)
$\text{Standard deviation} = \sqrt{\text{npq}}$
$=\sqrt{400\times\frac{1}{10}\times\frac{9}{10}}$
$=\sqrt{36}$
$\text{standard deviation}=6$
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Question 193 Marks
A die is tossed twice. Find the probability of getting a number greater than 3 on each toss.
Answer
Given, A and B are independent events and $\text{P}(\text{A}\cap\text{B})=0.60,\text{P(A)}=0.2$
A and B are independent events,
So, $\text{P}(\text{A}\cap\text{B})=\text{P(A) }\text{P(B)}$
We know that,
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(A)}-\text{P}(\text{A}\cap\text{B})$
$0.6=0.2+\text{P(B)}-\text{P(B) }\text{P(B)}$
$0.6-0.2=\text{P(B)}-0.2\text{P(B)}$
$0.4=0.8\text{P(B)}$
$\text{P(B)}=\frac{0.4}{0.8}$
$\text{P(B)}=0.5$
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Question 203 Marks
An instructor has a test bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the test bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Answer
Number of easy True/False questions = 300
Number of difficult True/False questions = 200
Number of easy multiple choice questions = 500
Number of difficult multiple choice questions = 400
Total number of all such questions = n(S) = 1400
Let E represents an easy question and F represents a multiple choice question.
$\therefore\ \text{n}(\text{E})\ =300+500=800$
$\text{and}\ \text{n}(\text{F})\ =500+400=900$
$\text{P}(\text{F})=\frac{\text{n}(\text{F})}{\text{n}(\text{S})}=\frac{900}{1400}$
$\text{n}(\text{E}\cap\text{F})=500\ \Rightarrow\ \ \ \ \ \text{P}(\text{E}\cap\text{F})=\frac{\text{n}(\text{E}\cap\text{F})}{\text{n}(\text{S})}=\frac{500}{1400}$
$\text{P}(\text{E}|\text{F})=\frac{\text{P}(\text{E}\cap\text{F})}{\text{P}(\text{F})}=\frac{\frac{500}{1400}}{\frac{900}{1400}}=\frac{500}{900}=\frac{5}{9}$
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Question 213 Marks
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?
Answer
When a die is thrown, the sample space (S) is$\text{S}=\{1,\ 2,\ 3,\ 4,\ 5,\ 6\}$
Let A: the number is even = {2, 4, 6}
$\Rightarrow\text{P}(\text{A})=\frac{3}{6}=\frac{1}{2}$
B: the number is red = {1, 2, 3}
$\Rightarrow\text{P}(\text{B})=\frac{3}{6}=\frac{1}{3}$
$\therefore\text{A}\cap\text{B}=\left\{2\right\}$
$\text{P}(\text{AB})=\text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
$\text{P}(\text{A}).\text{P}(\text{B})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\neq\frac{1}{6}$
$\Rightarrow\text{P}(\text{A}).\text{P}(\text{B})\neq\text{P}(\text{AB})$
Therefore, A and B are not independent.
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Question 223 Marks
Determine P(E|F) : A coin is tossed three times.
E : heads on third toss, F : heads on first two tosses.
Answer
A coin tossed three times, i.e.,$\text{S}=(\text{TTT, HTT, THT, TTH, HHT, HTH, THH, HHH})$$\ \ \ \ \ \Rightarrow\ \ \ \ \ \ \ \text{n}(\text{S})=8$
E : heads on third toss
$\text{E}=(\text{TTH, HTH, THH, HHH})$$\ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \ \text{n}(\text{E})=4$
$\text{P}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{4}{8}=\frac{1}{2}$
F : heads on first two tosses
$\text{F}(\text{HHT, HHH})$$\ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \ \ \ \ \text{n}(\text{F})=2$
$ \text{P}\left(\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{2}{8}=\frac{1}{4}$
$\therefore\ \ \ \ \ \ \ \ \text{E}\cap\text{F}=\left(\text{HHH}\right)\ \ \ \ \ \ \Rightarrow\text{n}\left(\text{E}\cap\text{F}\right)=1$
$\therefore\ \ \ \ \ \ \ \ \text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\cap\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{1}{8}$
$\text{And}\ \ \ \ \text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\cap\text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac{1}{8}}{\frac{2}{8}}=\frac{1}{2}$
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Question 233 Marks
Find the mean and standard deviation of the following probability distributions:
$\text{x}_\text{i}$
$0$
$1$
$2$
$3$
$4$
$5$
$\text{p}_\text{i}$
$\frac{1}{6}$
$\frac{5}{18}$
$\frac{2}{9}$
$\frac{1}{6}$
$\frac{1}{9}$
$\frac{1}{18}$
Answer
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}^2$
$0$ $\frac{1}{6}$ $0$ $0$
$1$ $\frac{5}{18}$ $\frac{5}{18}$ $\frac{5}{18}$
$2$ $\frac{2}{9}$ $\frac{4}{9}$ $\frac{8}{9}$
$3$ $\frac{1}{6}$ $\frac{1}{2}$ $\frac{3}{2}$
$4$ $\frac{1}{9}$ $\frac{4}{9}$ $\frac{16}{9}$
$5$ $\frac{1}{18}$ $\frac{5}{18}$ $\frac{25}{18}$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{35}{18}$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{35}{6}$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{35}{18}$
Variance $\sum\text{p}_\text{i}\text{x}_\text{i}^2-\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2=\frac{35}{6}-\Big(\frac{35}{18}\Big)^2=\frac{665}{324}$
Standard Deviation $=\sqrt{\text{Variation}}=\frac{\sqrt{665}}{18}$
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Question 243 Marks
Find the mean and standard deviation of the following probability distributions$:$
$x_i$ $-2$ $-1$ $0$ $1$ $2$
$p_i$ $0.1$ $0.2$ $0.4$ $0.2$ $0.1$
Answer
$x_i$ $p_i$ $x_ip_i$ $x_i^2p_i$
$-2$ $0.1$ $-0.2$ $0.4$
$-1$ $0.2$ $-0.2$ $0.2$
$0$ $0.4$ $0$ $0$
$1$ $0.2$ $0.2$ $0.2$
$2$ $0.1$ $0.2$ $0.4$
    $\sum\text{xp}=0$ $\sum\text{x}^2\text{p=1.2}$
Mean $=\sum\text{xp}$
Mean $=0$
Standard Deviation $=\sqrt{\sum\text{x}^2\text{p}-(\text{mean})^2}$
$=\sqrt{(1.2)^2-(0)^2}$
$=1.095$
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Question 253 Marks
Find the mean and standard deviation of the following probability distributions:
$\text{x}_\text{i}$
$-5$
$-4$
$1$
$2$
$\text{p}_\text{i}$
$\frac{1}{4}$
$\frac{1}{8}$
$\frac{1}{2}$
$\frac{1}{8}$
Answer
$\text{x}_\text{i}$ $\text{p}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}$ $\text{p}_\text{i}\text{x}_\text{i}^2$
$-5$ $\frac{1}{4}$ $-\frac{5}{4}$ $\frac{25}{4}$
$-4$ $\frac{1}{8}$ $-\frac{4}{8}$ $\frac{16}{8}$
$1$ $\frac{1}{2}$ $\frac{1}{2}$ $\frac{1}{2}$
$2$ $\frac{1}{8}$ $\frac{2}{8}$ $\frac{4}{8}$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=-1$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{74}{8}$
$\sum\text{p}_\text{i}\text{x}_\text{i}=-1$
Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$
$=\frac{74}{8}-(-1)^2$
$=9.25-1$
$=8.25$
Step Deviation $=\sqrt{\text{Variance}}$
$=\sqrt{8.25}$
$=2.872$
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Question 263 Marks
Find the mean and standard deviation of the following probability distributions$:$
$x_i$ $-3$ $-1$ $0$ $1$ $3$
$p_i$ $0.05$ $0.45$ $0.20$ $0.25$ $0.05$
Answer
$x_i$ $p_i$ $x_ip_i$ $x_i^2p_i$
$-3$ $0.05$ $-0.15$ $0.45$
$-1$ $0.45$ $-0.45$ $0.45$
$0$ $0.20$ $0$ $0$
$1$ $0.25$ $0.25$ $0.25$
$3$ $0.05$ $0.15$ $0.45$
    $\sum\text{xp}=-0.2$ $\sum\text{x}^2\text{p=1.6}$
Mean $=\sum\text{xp}$
Mean $=0.2$
Standard Deviation $=\sqrt{\sum\text{x}^2\text{p}-(\text{mean})^2}$
$=\sqrt{(1.6)-(-0.2)^2}$
$=\sqrt{1.6-0.04}$
$=\sqrt{1.56}$
Standard Deviation $=1.249$
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Question 273 Marks
Find the mean and standard deviation of the following probability distributions$:$
$x_i$ $2$ $3$ $4$
$p_i$ $0.2$ $0.5$ $0.3$
Answer
$x_i$ $p_i$ $p_ix_i$ $p_ix_i^2$
$2$ $0.2$ $0.4$ $0.8$
$3$ $0.5$ $1.5$ $4.5$
$4$ $0.3$ $1.2$ $4.8$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=3.1$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=10.1$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=3.1$
Variation $=\sum\text{p}_\text{i}\text{x}_\text{i}^2$
$=\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2$
$=10.1-(3.1)^2$
$=0.49$
Standard deviation $=\sqrt{\text{Variance}}=0.7$
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Question 283 Marks
Find the mean and standard deviation of the following probability distributions$:$
$x_i$ $-1$ $0$ $1$ $2$ $3$
$p_i$ $0.3$ $0.1$ $0.1$ $0.3$ $0.2$
Answer
$x_i$ $p_i$ $p_ix_i$ $p_ix_i{}^2$
$-1$ $0.3$ $-0.3$ $0.3$
$0$ $0.1$ $0$ $0$
$1$ $0.1$ $0.1$ $0.1$
$2$ $0.3$ $0.6$ $0.2$
$3$ $0.2$ $0.6$ $1.8$
    $\sum\text{p}_\text{i}\text{x}_\text{i}=1$ $\sum\text{p}_\text{i}\text{x}_\text{i}^2=3.4$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=1$
Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-(\text{Mean})^2$
$3.4-1$
$=2.4$
Step Deviation $=\sqrt{\text{Variance}}$
$=\sqrt{2.4}$
$=1.549$
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Question 293 Marks
If in a binomial distribution n = 4 and P(X = 0) $=\frac{16}{81},$ find q.
Answer
Given that,
$\text{n}=4,\text{P(X}=0)=\frac{16}{81}$
We know that,
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X = r})=\text{ }^4\text{c}_{\text{r}}(\text{p})^{\text{r}}(\text{q})^{4-\text{r}}$
$\text{P(X}=0)=\text{ }^4\text{c}_0(1-\text{q})^0(\text{q})^{4-0}$
$\frac{16}{81}=1.1.\text{q}^4$
$\text{q}^4=\big(\frac{2}{3}\big)^4$
$\text{q}=\frac{2}{3}$
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Question 303 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 $\frac{44}{91}.$
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Question 313 Marks
A couple has two children,
  1. Find the probability that both children are males, if it is known that at least one of the children is male.
  2. Find the probability that both children are females, if it is known that the elder child is a female.
Answer
  1. Let $B_1$ and $g_1$ stand for male and female respectively.
Now the sample space is $(S) =\{B_1B_2, B_1G_2, B_2G_1, G_1G_2\}$
Let us consider the following events,
$A =$ both are males
$B =$ at least one is a male
$\therefore A = \{B_1B_2\}$ and $B = \{B_1B_2, B_1G_2, B_2G_1\}$
$\text{P}(\text{B})=\frac{3}{4},\ \text{A}\cap\text{B}=\{\text{B}_1\text{B}_2\}\ \text{and}\ \text{P}(\text{A}\cap\text{B})=\frac{1}{4}$
$\therefore$ Required probability $=\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$
  1. Let $A =$ both are females $A =\{G_1G_2\}$ and $C =$ the older is a girl
$\text{C}=\{\text{B}_1\text{B}_2,\ \text{B}_1\text{G}_2\}\ \Rightarrow\ \text{P}(\text{C})=\frac{2}{4}$
$\therefore$ Required probability $=\text{P}(\text{A}|\text{C})=\frac{\text{P}(\text{A}\cap\text{C})}{\text{P}(\text{C})}=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2}$
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Question 323 Marks
Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
Answer
$\text{Given}:\ \ \text{P}(\text{G1})=0.6,\ \text{P}(\text{G2})=0.4$Let P denotes the launching of new product.
$\therefore\ \text{P}(\text{PIG}_1)=0.7,\ \text{P}(\text{PIG}_2)=0.3$
$ \text{P}(\text{G}_1|\text{P})=\frac{\text{P}(\text{G}_2)\text{P}(\text{P}|\text{G}_2)}{\text{P}(\text{G}_1)\text{P}(\text{P}|\text{G}_1)+\text{P}(\text{G}_2)\text{P}(\text{P}|\text{G}_2)}$
$=\frac{0.4\times0.3}{0.6\times0.7+0.4\times0.3}=\frac{12}{54}=\frac{2}{9} $
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Question 333 Marks
Three events A, B and C have probabilities $\frac{2}{5},\frac{1}{3}$ and $\frac{1}{2},$ respectively. Given than $\text{P}(\text{A}\cap\text{C})=\frac{1}{5}$ and $\text{P}(\text{B}\cap\text{C})=\frac{1}{4},$ find the values of $\text{P}\Big(\frac{\text{C}}{\text{B}}\Big)$ and $\text{P}(\text{A}'\cap\text{C}').$
Answer
Here, $\text{P}(\text{A})=\frac{2}{5},\text{P}(\text{B})=\frac{1}{3},\text{P}(\text{C})=\frac{1}{2},\text{P}(\text{B}\cap\text{C})=\frac{1}{4}$ and $\text{P}(\text{B}\cap\text{C})=\frac{1}{4}$
$\therefore\text{P}\Big(\frac{\text{C}}{\text{B}}\Big)=\frac{\text{P}(\text{B}\cap\text{C})}{\text{P}(\text{B})}=\frac{\frac{1}{4}}{\frac{1}{3}}=\frac{3}{4}$
And $\text{P}(\text{A}'\cap\text{C}')=1-\text{P}(\text{A}\cup\text{C})$
$=1-\big[\text{P}(\text{A})+\text{P}(\text{C})-\text{P}(\text{A}\cap\text{C})\big]$
$=1-\Big[\frac{2}{5}+\frac{1}{2}-\frac{1}{5}\Big]=1-\Big[\frac{4+5-2}{10}\Big]$
$=1-\frac{7}{10}=\frac{3}{10}$
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Question 343 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 $=\sqrt{\text{Variance}}$
$=\sqrt{1}$
$=1$
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Question 353 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)
$=\frac{\text{}^{5}\text{C}_4}{\text{}^{25}\text{C}_4}$
$=\frac{5}{12650}$
$=\frac{1}{2530}$
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Question 363 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 $\text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{A}\cup\text{B})=\frac{3}{5},\ \text{and}\ \text{P}(\text{B})=\text{p}$
  1. 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, $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\frac{3}{5}=\frac{1}{2}+\text{p}-0$
$\Rightarrow\ \text{p}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$
  1. 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, $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \frac{3}{5}=\frac{1}{2}+\text{p}-\frac{1}{2}\text{p}$
$\Rightarrow\frac{3}{5}=\frac{1}{2}+\frac{\text{p}}{2}$
$\Rightarrow\ \frac{\text{p}}{2}=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$
$\Rightarrow\text{p}=\frac{2}{10}=\frac{1}{5}$
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Question 373 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 $=\sqrt{\text{Variance}}$
$=\sqrt{2.24}$
$=1.497$
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Question 383 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 $= 1.732$
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Question 393 Marks
An urn contains 3 white, 4 red and 5 black balls. Two balls are drawn one by one without replacement. What is the probability that at least one ball is black?
Answer
Urn contains 3 white, 4 red and 5 black balls. Total balls = 12
Two balls are drawn without replacement
A = First ball is black
B = Second ball is black
P (Atleast one ball is black)
$=\text{P}(\overline{\text{A}}\cup\overline{\text{B}})$
$=1-\text{P}(\overline{\text{A}}\cup\overline{\text{B}})$
$=1-\text{P}(\overline{\text{A}})\text{P }\Big(\overline{\frac{\text{B}}{\text{A}}}\Big)$
$=1-\Big(\frac{7}{12}\times\frac{6}{12}\Big)$
$=1-\frac{7}{22}$
Required probability $=\frac{15}{22}$
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Question 403 Marks
A, B, and C are independent witness of an event which is known to have occurred. A speaks the truth three times out of four, B four times out of five and C five times out of six. What is the probability that the occurrence will be reported truthfully by majority of three witnesses?
Answer
P(A speaks truth) $=\frac{3}{4}$
P(B speaks truth) $=\frac{4}{5}$
P(C speaks truth) $=\frac{5}{6}$
P(majority speaks truth) = P(two speaks truth) + P(all speak truth)
= P(A) × P(B)[1 - P(C)] + P(A) × P(C)[1 - P(B)] + P(C) × P(B)[1 - P(A)] + P(A) × P(B) × P(C)
$=\frac{3}{4}\times\frac{4}{5}\Big(1-\frac{5}{6}\Big)+\frac{3}{4}\times\frac{5}{6}\Big(1-\frac{4}{5}\Big) \\ +\frac{4}{5}\times\frac{5}{6}\Big(1-\frac{3}{4}\Big)+\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}$
$=\frac{12}{120}+\frac{15}{120}+\frac{20}{120}+\frac{60}{120}$
$=\frac{107}{120}$
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Question 413 Marks
The probability of a man hitting a target is 1/4. If he fires 7 times, what is the probability of his hitting the target at least twice?
Answer
Let X be number of times the target is hit. Then, X follows a binomial distribution with n = 7,
$\text{p}=\frac{1}{4}$ and $\text{q}=\frac{3}{4}$
$\text{P}(\text{X = r})=\text{ }^7\text{C}_{\text{r}}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{7-\text{r}}$
P (hitting the target at least twice)
$=\text{P}(\text{X}\geq2)$
$=1-\big\{\text{P}(\text{X}=0)+\text{P}(\text{X}=1)\big\}$
$=1-\text{ }^7\text{C}_0\big(\frac{1}{4}\big)^0\big(\frac{3}{4}\big)^{7-0}-\text{ }^7\text{C}_1\big(\frac{1}{4}\big)^1\big(\frac{3}{4}\big)^{7-1}$
$=1-\big(\frac{3}{4}\big)^7-7\big(\frac{1}{4}\big)\big(\frac{3}{4}\big)^6$
$=1-\frac{1}{16384}(2187+5103)$
$=1-\frac{3645}{8192}$
$=\frac{4547}{8192}$
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Question 423 Marks
Eight coins are thrown simultaneously. Find the chance of obtaining at least six heads.
Answer
Let X be the number of heads in tossing 8 coins.
X follows a binomial distribution with $\text{n}=8;\text{p}=\frac{1}{2}$ and $\text{q}=\frac{1}{2};$
$\text{P}(\text{X = r})=\text{ }^8\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{8-\text{r}}=\text{ }^8\text{C}_{\text{r}}\big(\frac{1}{2}\big)^8$
Probability of obtaining at least 6 heads $=\text{P}(\text{X}\geq6)$
$=\text{P}(\text{X}=6)+\text{P}(\text{X}=7)+\text{P}(\text{X}=8)$
$=\text{ }^8\text{C}_6\big(\frac{1}{2}\big)^8+\text{ }^8\text{C}_7\big(\frac{1}{2}\big)^8+\text{ }^8\text{C}_8\big(\frac{1}{2}\big)^8$
$=\frac{1}{2^8}(28+8+1)$
$=\frac{37}{256}$
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Question 433 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, $\text{n}=8$
$\text{Mean = np}$
$=\frac{8}{3}$
$=2.66$
$\text{Standard deviation}=\sqrt{\text{npq}}$
$=\sqrt{8\times\frac{1}{3}\times\frac{2}{3}}$
$=\frac{4}{3}$
$=1.33$
$\text{Mean}=2.66,\text{Standard deviation}=1.33$
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Question 443 Marks
An unbiased die is thrown twice. A success is getting a number greater than 4. Find the probability distribution of the number of successes.
Answer
Let X denote getting a number greater than 4.Then, X follow a binomial distribution with n =2
$\text{p}=\text{P(X > 4)}=\text{P}\text{(X = 5 or 6)}$
$=\frac{1}{6}+\frac{1}{6}$
$=\frac{1}{3}$
$\text{q}=1-\text{p}=\frac{2}{3}$
$\text{P}(\text{X = r})=\text{ }^2\text{C}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{2-\text{r}},\text{r}=0, 1, 2, $
Substituting for r we get probability distribution of X as follows.
$\text{X}$ $0$ $1$ $2$
$\text{P(X)}$ $\frac{4}{9}$ $\frac{4}{9}$ $\frac{1}{9}$
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Question 453 Marks
A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?
Answer
$\text{Given:}\ \ \ \text{P}(\text{A})=\frac{60}{100},\ \text{P}(\text{B})=\frac{40}{100}$Let D denotes a defective item:
$\therefore\ \text{P}(\text{D}|\text{A})=\frac{2}{100}\ \text{and}\ \text{P}(\text{D}|\text{B})=\frac{1}{100}$
$\text{P}(\text{B}|\text{D})=\frac{\text{P}(\text{B})\text{P}(\text{D}|\text{B})}{\text{P}(\text{A})\text{P}(\text{D}|\text{A})+\text{P}(\text{B})\text{P}(\text{D}|\text{B})}$
$=\frac{\frac{40}{100}\times\frac{1}{100}}{\frac{60}{100}\times\frac{2}{100}+\frac{40}{100}\times\frac{1}{100}}=\frac{40}{120+40}=\frac{40}{160}=\frac{1}{4}$
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Question 463 Marks
Let $X$ be a random variable which assumes values $x_1, x_2, x_3, x_4$ such that $2P(X = x_1) = 3P(X = x_2) = P(X = x_3) = 5P(X = x_4).$ 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,
$\text{X}:$
$\text{x}_1$
$\text{x}_2$
$\text{x}_3$
$\text{x}_4$
$\text{P}(\text{X}):$
$\frac{15}{61}$
$\frac{10}{61}$
$\frac{30}{61}$
$\frac{6}{61}$
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Question 473 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 $=\sum\text{x}_\text{i}\text{p}_\text{i}^2-(\text{Mean})^2$
$=60.7-49$
$=11.7$
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Question 483 Marks
On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?
Answer
The repeated guessing of correct answers from multiple choice questions are Bernoulli trials. Let X represent the number of correct answers by guessing in the set of 5 multiple choice questions.
Probability of getting a correct answer is, $\text{p}=\frac{1}{3}$
$\therefore\ \text{q}=1-\text{p}=1-\frac{1}{3}=\frac{2}{3}$
Clearly, X has a binomial distribution with n = 5 and $\text{p}=\frac{1}{3}$
$\therefore\ \text{P}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x}$
$=\ ^5\text{C}_\text{x}\bigg(\frac{2}{3}\bigg)^{5-\text{x}}.\bigg(\frac{1}{3}\bigg)^\text{x}$
P(guessing more than 4 correct answers) = P(X ≥ 4)
$=\text{P}(\text{X}=4)+\text{P}(\text{X}=5)$
$=\ ^5\text{C}_\text{4}\bigg(\frac{2}{3}\bigg).\bigg(\frac{1}{3}\bigg)^4+\ ^5\text{C}_\text{5}\bigg(\frac{1}{3}\bigg)^5$
$=5\cdot\frac{2}{3}\cdot\frac{1}{81}+1\cdot\frac{1}{243}$
$=\frac{10}{243}+\frac{1}{243}$
$=\frac{11}{243}$
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Question 493 Marks
A dice is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?
Answer
The sample space of the experiment is {(1, 1), (1, 2), (1, 3), ....., (6, 6)} conisting of 36 outcomes.
$\text{P(A)}=\text{P}(\text{Sum}=6)=\frac{5}{36}$
$\text{P(B)}=\text{P}(4\text{ appears at least once})=\frac{11}{36}$
Now, $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A and B})}{\text{P(A)}}$
$=\frac{\text{P}(\text{Sum is } 6 \text{ and }4 \text{ has appeard at least once})}{\text{P(A)}}$
$=\frac{\frac{2}{36}}{\frac{5}{36}}$
$=\frac{2}{5}$
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Question 503 Marks
If the sum of the mean and variance of a binomial distribution for 6 trials is $\frac{10}{9},$ find the distribution.
Answer
Given that $\text{n}=6$
The sum of mean and variance of a binomial distribution for 6 trials is $\frac{10}{3}.$
$\Rightarrow6\text{p}+6\text{pq}=\frac{10}{3}$
$\Rightarrow18\text{p}+18\text{p}(1-\text{p})=10$
$\Rightarrow18\text{p}^2-36\text{p}+10=0$
$\Rightarrow(3\text{p}-1)(6\text{p}-10)=0$
$\Rightarrow\text{p}=\frac{1}{3}$ or $\frac{5}{3}$
$\text{p}=\frac{5}{3}$ (neglected as it is greater than 1)
$\therefore\text{p}=\frac{1}{3}$
$\Rightarrow\text{q}=1-\text{p}=\frac{2}{3}$
Hence, the distribution is given by
$\text{P(X = r})\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{6-\text{r}},\text{r}=0,1,2\dots6$
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