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Maths 3 Marks Question

Binomial Distribution · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 42 questions.

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Question 13 Marks
There are 6% defective items in a large bulk of items. Find the probability that a sample of 8 items will include not more than one defective item.
Answer
Let x denote the number of defective items in a sampla of 8 items. athen, x follows a binomial distribution with n = 8,
P = (Probability of getting a defective item) = 0.06 and $q = 1 - P = 0.94$
$P(X = r) = {^8C_r}(0.06)^r(0.94)^{8-r}, r = 0, 1, 2, 3, .....8$
The required probability = probebility of not more than one defective item
$=\text{P}(\text{X}\leq1)$
$=\text{P}(\text{X}=0)+\text{P}(\text{x}=1)$
$=\text{ }^8\text{C}_0 (0.06)^0(0.94)^{8-0}+\text{ }^8\text{C}_1(0.06)^1(0.94)^{8-1}$
$=(0.94)^8+8(0.06)(0.94)^7$
$=(0.94)^7\big\{0.94+0.48\big\}$
$=1.42(0.94)^7$
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Question 23 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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Question 33 Marks
Can the mean of a binomial distribution be less than its variance?
Answer
Let X be a binomial veriate with parameters n and p.
Mean = np varience = npq Mean - variance $= np - npq = np (1 - q) = np.$
$p = np^2$ Mean - variance > 0 Mean > variance
So, mean can never be less than variance.
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Question 43 Marks
A man wins a rupee for head and loses a rupee for tail when a coin is tossed. Suppose that he tosses once and quits if he wins but tries once more if he loses on the first toss. Find the probability distribution of the number of rupees the man wins.
Answer
Let n denote the number of throws required to get a head and X denote the amount won / lost.
He may get head on first toss or lose first and $2^{nd}$ toss or lose and won second toss probability distribution for X
$\text{Number of throws (n)}:$ $1$ $2$ $2$
$\text{Amount won/lost(x)}:$ $1$ $0$ $-2$
$\text{Probability P (X)}:$ $\frac{1}{2}$ $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$ $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$
So probability distribution is given by
$\text{X}$ $\text{P(X)}$
$0$ $\frac{1}{4}$
$1$ $\frac{1}{2}$
$-2$ $\frac{1}{4}$
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Question 53 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 63 Marks
A coin is tossed 5 times. What is the probability that tail appears an odd number of times?
Answer
Let X denote the number of tails when a coin is tossed 5 times.
X follows a binomial distribution with $\text{n}=5;\text{p}=\frac{1}{2};\text{q}=1-\text{p}=\frac{1}{2}$
Then $\text{P}(\text{X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{\text{n}-\text{r}}=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^5$
The required probability $=\text{P}(\text{X = odd})$
$=\text{P}(\text{X}=1)+\text{P}(\text{X}=3)+\text{P}(\text{X}=5)$
$=\text{ }^5\text{C}_1\big(\frac{1}{2}\big)^5+\text{ }^5\text{C}_3\big(\frac{1}{2}\big)^5+\text{ }^5\text{C}_5\big(\frac{1}{2}\big)^5$
$=\big(\frac{1}{2}\big)^5\ [5+10+1]$
$=\frac{16}{32}$
$=\frac{1}{2}$
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Question 73 Marks
Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of spades. Hence, find the mean of the distribution.
Answer
We have,
p = probability of getting a spade in a draw $=\frac{13}{52}=\frac{1}{4}$ and $\text{q}=1-\text{p}=1-\frac{1}{4}=\frac{3}{4}$
Let X denote a success of getting a spade in a throw. Then,
X follows binomial distribution with parameters $\text{n}=3$ and $\text{p}=\frac{1}{4}$
$\therefore\text{P(X = r})=\text{ }^3\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(3-\text{r})}=\text{ }^3\text{c}_{\text{r}}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{(3-\text{r})}=\frac{\text{ }^{3}\text{c}_{\text{r}}3^{(3-\text{r})}}{4^3}=\frac{27}{64}\Big(\frac{\text{ }^3\text{c}_{\text{r}}}{3^{\text{r}}}\Big),$ where, $\text{r}=0,1,2,3$
So, the probability distibution of X is given by:
$\text{P(X = r})=\frac{27}{64}\Big(\frac{\text{ }^3\text{C}_{\text{r}}}{3^{\text{r}}}\Big),$ where, $\text{r}=0,1,2,3$
Now,
$\text{Mean, E(X) = np}=3\times\frac{1}{4}=\frac{3}{4}$
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Question 83 Marks
A pair of dice is thrown 6 times. If getting a total of 9 is considered a success, what is the probability of at least 5 successes?
Answer
Let X be the number of successes in 6 throws of the two dice. Probability of success = Probability of getting a total of 9 = Probability of getting (3, 6), (4, 5), (5, 4), (6, 3) out of 36 outcomes $\text{p}=\frac{4}{36}=\frac{1}{9},\text{q}=1-\text{p}=\frac{8}{9}\text{ and }\text{n}=6$ X follows a binomial distribution with $\text{n}=6,\text{p}=\frac{1}{9}$ and $\text{q}=\frac{8}{9}$$\text{P}(\text{X = r})=\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{9}\big)^{\text{r}}\frac{8}{9}^{6-\text{r}}$
The required probability = Probability of at least 5 success $=\text{P}(\text{X}\geq5)$ $=\text{P}(\text{X}=5)+\text{P}(\text{X}=6)$ $=\text{ }^6\text{C}_56\big(\frac{1}{9}\big)^5\big(\frac{8}{9}\big)^{6-5}+\text{ }^6\text{C}_6\big(\frac{1}{9}\big)^6\big(\frac{8}{9}\big)^{6-6}$ $=\frac{6(8)+1}{9^6}=\frac{49}{9^6}$
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Question 93 Marks
A box has 20 pens of which 2 are defective. Calculate the probability that out of 5 pens drawn one by one with replacement, at most 2 are defective.
Answer
The probability of getting a defective pen from the box is $\frac{2}{20}=\frac{1}{10}$
out of the five pens the probability of getting at most 2 defective pens is $\frac{1}{5}$ or $\frac{2}{5}$
The probability of getting a defective pen from the five picked is thus:
$\big(\frac{1}{5}+\frac{2}{5}\big)\times\frac{1}{10}=\frac{3}{5}\times\frac{1}{10}=\frac{3}{50}$
$\frac{3}{50}$
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Question 103 Marks
Find the probability that in 10 throws of a fair die, a score which is a multiple of 3 will be obtained in at least 8 of the throws.
Answer
Here success is a score which is multiple of 3 i.e. 3 or 6.
$\therefore\text{p}(3 \text{ or } 6)=\frac{2}{6}=\frac{1}{3}$
The probability of r success in 10 throws is given by
$\text{P(r)}=\text{ }^{10}\text{C}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{10-\text{r}}$
Now P(at least 8 success) =P(8) + P(9) + P(10)
$=\text{ }^{10}\text{C}_8\big(\frac{1}{3}\big)^8\big(\frac{2}{3}\big)^2+\text{ }^{10}\text{C}_9\big(\frac{1}{3}\big)^9\big(\frac{2}{3}\big)^1+\text{ }^{10}\text{C}_{10}\big(\frac{1}{3}\big)^{10}\big(\frac{2}{3}\big)^{0}$
$=\frac{1}{3^{10}}\big[45\times4+10\times2+1\big]$
$=\frac{20^1}{3^{10}}$
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Question 113 Marks
An urn contains four white and three red balls. Find the probability distribution of the number of red balls in three draws with replacement from the urn.
Answer
As three balls are drawn with replacement, the number of white balls, say X, follows binomial distribution with n = 3
$\text{p}=\frac{3}{7}$ and $\text{q}=\frac{4}{7}$
$\text{P}(\text{X = r})=\text{ }^{3}\text{C}_{\text{r}}\big(\frac{3}{7}\big)^{\text{r}}\big(\frac{4}{7}\big)^{3-\text{r}},\text{r}=0,1,2,3$
$\text{P}(\text{X}=0)=\text{ }^{3}\text{C}_0\big(\frac{3}{7}\big)^0\big(\frac{4}{7}\big)^{3-0}$
$\text{P}(\text{X}=1)=\text{ }^3\text{C}_1\big(\frac{3}{7}\big)^1\big(\frac{4}{7}\big)^{3-1}$
$\text{P}(\text{X}=2)=\text{ }^2\text{C}_2\big(\frac{3}{7}\big)^2\big(\frac{4}{7}\big)^{3-2}$
$\text{P}(\text{X}=3)=\text{ }^3\text{C}_3\big(\frac{3}{7}\big)^3\big(\frac{4}{7}\big)^{3-3}$
$\text{X}$ $0$ $1$ $2$ $3$
$\text{P(X)}$ $\frac{64}{343}$ $\frac{144}{343}$ $\frac{108}{343}$ $\frac{27}{343}$
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Question 123 Marks
If the mean of a binomial distribution is 20 and its standard deviation is 4, find p.
Answer
let n and p be the parameter of binomial distribution. So
Given,
$\text{Mean = np}=20\dots(1)$
$\text{Standard deviation } =\sqrt{\text{npq}}=4$
Squaring both the sides,
$\text{npq}=16\dots(2)$
Dividing equation (2) by (1),
$\frac{\text{npq}}{\text{np}}=\frac{16}{20}$
$\text{q}=\frac{4}{5}$
$\text{p}=1-\text{q}$ [Since p + q = 1]
$=1-\frac{4}{5}$
$\text{p}=\frac{1}{5}$
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Question 133 Marks
Five dice are thrown simultaneously. If the occurrence of 3, 4 or 5 in a single die is considered a success, find the probability of at least 3 successes.
Answer
Let X denote the occurrence of 3, 4 or 5 in a single die. Then, X follows binomial distribution with n = 5.
Let p = probability of getting 3, 4 or 5 in a single die
$\text{p}=\frac{3}{6}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}=\frac{1}{2}$
$\text{P}(\text{X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}2{}\big)^{5-\text{r}}$
P(at least 3 successes) $=\text{P}(\text{X}\geq3)$
$=\text{P(X = 3)}+\text{P}(\text{X}=4)+\text{P}(\text{X}=5)$
$=\text{ }^5\text{C}_3\big(\frac{1}{2}\big)^3\big(\frac{1}{2}\big)^{5-3}+\text{ }^5\text{C}_4\big(\frac{1}{2}\big)^4\big(\frac{1}{2}\big)^{5-4}+\text{ }^5\text{C}_5\big(\frac{1}{2}\big)^5\big(\frac{1}{2}\big)^{5-5}$
$=\frac{\text{ }^5\text{C}_3+\text{ }^5\text{C}_4+\text{ }^5\text{C}_5}{2^5}$
$=\frac{1}{2}$
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Question 143 Marks
A coin is tossed 5 times. If X is the number of heads observed, find the probability distribution of X.
Answer
Let X = number of heads in 5 tosses. Then the binomial distribution
for X has $\text{n}=5,\text{p}=\frac{1}{2}$ and $\text{q}=\frac{1}{2}$
$\text{P(X = r)}=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{5-\text{r}},\text{r}=0, 1, 2, 3, 4, 5$
$=\frac{\text{ }^5\text{C}_{\text{r}}}{2^5}$
Substituting r = 0, 1, 2, 3, 4, 5 we get the following probability disrtribution.
$\text{X}$ $0$ $1$ $2$ $3$ $4$ $5$
$\text{P(X)}$ $\frac{1}{32}$ $\frac{5}{32}$ $\frac{10}{32}$ $\frac{10}{32}$ $\frac{5}{32}$ $\frac{1}{32}$
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Question 153 Marks
If the mean and variance of a binomial distribution are respectively 9 and 6, find the distribution.
Answer
Given: Mean = 9 and variance = 6
$\therefore\text{np}=9\dots(1)$
$\text{npq}=6\dots(2)$
Dividing eq. (2) by eq. (1), we get
$\text{q}=\frac{2}{3}$ and $\text{p}=1-\text{q}=\frac{1}{3}$
As np = 9, substituting the value of p, we get
$\frac{\text{n}}{3}=9$ or $\text{n}=27$
$\text{P(X = r})=\text{ }^{27}\text{C}_\text{r}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{27-\text{r}},\text{r}=0,1,2\dots27$
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Question 163 Marks
In a 20-question true-false examination, suppose a student tosses a fair coin to determine his answer to each question. For every head, he answers 'true' and for every tail, he answers 'false'. Find the probability that he answers at least 12 questions correctly.
Answer
Let p be the probability of answering a true. So
$\text{p}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}$ [Since p +q = 1]
$=\frac{1}{2}$
Thus the probability that he answers at least 12 questions correctly among 20 questions is
$\text{P(X}\geq12)=\text{P(X}=12)+\text{P(X}=13)+\text{P(X}=14)+\text{P(X}=15)\\+\text{P(X}=16)+\text{P(X}=17)+\text{P(X}=18)+\text{P(X}=19)+\text{P(X}=20)$
$$$=\big(\frac{1}{2}\big)^{20}\big\{\text{ }^{20}\text{C}_{12}+\text{ }^{20}\text{C}_{13}+\text{ }^{20}\text{C}_{14}+\text{ }^{20}\text{C}_{15}\\+\text{ }^{20}\text{C}_{16}+\text{ }^{20}\text{C}_{17}+\text{ }^{20}\text{C}_{18}+\text{ }^{20}\text{C}_{19}+\text{ }^{20}\text{C}_{20}\big\}$
$\frac{\text{ }^{20}\text{C}_{12}+\text{ }^{20}\text{C}_{13}+\text{ }^{20}\text{C}_{14}+\text{ }^{20}\text{C}_{15}+\text{ }^{20}\text{C}_{16}+\text{ }^{20}\text{C}_{17}+\text{ }^{20}\text{C}_{18}+\text{ }^{20}\text{C}_{19}+\text{ }^{20}\text{C}_{20}}{2^{20}}$
Therefore, the required answer is
$\frac{\text{ }^{20}\text{C}_{12}+\text{ }^{20}\text{C}_{13}+\text{ }^{20}\text{C}_{14}+\text{ }^{20}\text{C}_{15}+\text{ }^{20}\text{C}_{16}+\text{ }^{20}\text{C}_{17}+\text{ }^{20}\text{C}_{18}+\text{ }^{20}\text{C}_{19}+\text{ }^{20}\text{C}_{20}}{2^{20}}$
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Question 173 Marks
If the mean and variance of a binomial distribution are 4 and 3, respectively, find the probability of no success.
Answer
$\text{Mean (np)}=4$
$\text{Variance (npq)}=3$
$\Rightarrow\text{q}=\frac{3}{4}$
Hence, $\text{p}=1-\frac{3}{4}=\frac{1}{4}$
and $\text{n}=\frac{\text{Mean}}{\text{p}}=4\times4=16$
Therefore, the binomial distribution is given by
$\text{P(X = r})=\text{ }^{16}\text{C}_{\text{r}}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{16-\text{r}},\text{r}=0,1,2,\dots\text{r}$
Probability of no success $=\text{ }^{16}\text{C}_0\big(\frac{1}{4}\big)^{0}\big(\frac{3}{4}\big)^{16-0}=\big(\frac{3}{4}\big)^{16}$
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Question 183 Marks
If for a binomial distribution P(X = 1) = P(X = 2) = α, write P(X = 4) in terma of α.
Answer
For binomial distribution of X,
$\text{P(X = r})=\text{ }^{\text{n}}\text{C}_{\text{r}}(\text{p})^{\text{r}}(\text{q})^{\text{n}-\text{r}},\text{r}=0,1,2,\dots,\text{n}$
$\text{P(X}=1)=\text{np(q)}^{\text{n}-1}$
$\text{P(X}=2)=\text{ }^{\text{n}}\text{C}_2\text{p}^2(\text{q})^{\text{n}-2}$
$\Rightarrow\text{np(q)}^{\text{n}-1}=\text{ }^{\text{n}}\text{C}_2\text{p}^2(\text{q})^{\text{n}-2}=\alpha$
Simplifying the above equation we get,
$\text{q}=\frac{\text{n}-1}{2}\text{p}$
$\Rightarrow2\text{q}=\text{np}-\text{p}$
On putting, $\text{q}=1-\text{p}$ we get
$2-2\text{p}=\text{np}-\text{p}$
$\text{p(n+1})=2\dots(1)$
Also, $\text{P(X}=1)=\alpha$
$\Rightarrow\text{np}(1-\text{p})^{\text{n}-1}=\alpha\dots(2)$
Note: We cannot find the value of n as (1) and (2) are not linear and hence we cannot find the value of P(X = 4)
Alternate Answer
Binomial distribution $\text{ }^{\text{n}}\text{C}_{\text{x}}\text{p}^{\text{x}}\text{q}^{(\text{n}-\text{x})}$
let X be the discrete variable, n the sample size
$\text{P(X}=1)=\text{ }^{\text{n}}\text{C}_1\text{p}^1\text{q}^{(\text{n}-1)}$
$\text{P(X}=2)=\text{ }^{\text{n}}\text{C}_2\text{p}^2\text{q}^{(\text{n}-2)}$
Given $\text{P(X}=1)=\text{P(X}=2)=\alpha$
$\text{ }^{\text{n}}\text{C}_1\text{p}^1\text{q}^{(\text{n}-1)}=\text{ }^{\text{n}}\text{C}_2\text{p}^2\text{q}^{(\text{n}-2)}=\alpha$
$\text{npq}^{(\text{n}-1)}=\alpha\Rightarrow\text{q}^{\text{n}}=\frac{\alpha}{\text{n}}\times\frac{\text{q}}{\text{p}}$
$\text{P(X}=4)=\text{ }^{\text{n}}\text{C}_4\text{p}^4\text{q}^{(\text{n}-4)}=\text{ }^{\text{n}}\text{C}_4\text{p}^4\frac{\text{q}^{\text{n}}}{\text{q}^4}$
$=\text{ }^{\text{n}}\text{C}_4\text{p}^4\frac{1}{\text{q}^4}\times\frac{\alpha}{\text{n}}\times\frac{\text{q}}{\text{p}}$
$=\text{ }^{\text{n}}\text{C}_4\alpha\times\Big(\frac{\text{p}}{\text{q}}\Big)^3$
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Question 193 Marks
Assume that on an average one telephone number out of 15 called between 2 P.M. and 3 P.M. on week days is busy. What is the probability that if six randomly selected telephone numbers are called, at least 3 of them will be busy?
Answer
Let X be the number of busy calls for 6 randomly selected telephone nimbers.
X following a binomial distribution with $\text{n}=6;\text{p}=1$ out of $15=\frac{1}{15}$ and $\text{q}=\frac{14}{15}$
$\text{p}(\text{X = r})=\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{15}\big)^{\text{r}}\big(\frac{14}{15}\big)^{6-\text{r}}$
Probability that at least 3 of them are busy $=\text{P}(\text{X}\geq3)$
$=1-\big\{\text{P}(\text{X}=0)+\text{P}(\text{X}=1)+\text{P}(\text{X}=2)\big\}$
$=1-\big\{\text{ }^6\text{C}_0\big(\frac{1}{15}\big)^0\big(\frac{14}{15}\big)^{6-0}+\text{ }^6\text{C}_1\big(\frac{1}{15}\big)^1\big(\frac{14}{15}\big)^{6-1}+\text{ }^6\text{C}_2\big(\frac{1}{15}\big)^2\big(\frac{14}{15}\big)^{6-2}\big\}$
$=1-\big\{\big(\frac{14}{15}\big)^6+\frac{6}{15}\big(\frac{14}{15}\big)^5+\frac{1}{15}\big(\frac{14}{15}\big)^4\big\}$
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Question 203 Marks
A dice is thrown thrice. A success is 1 or 6 in a throw. Find the mean and variance of the number of successes.
Answer
Let p be the probability of success, so
$\text{p}=\frac{2}{6}$ [Since success in occurance of 1 or 6 on the die]
$\text{p}=\frac{1}{3}$
Given, $\text{n}=3$
$\text{q}=1-\text{p}$ [Since p + q = 1]
$=1-\frac{1}{3}$
$\text{q}=\frac{2}{3}$
$\text{Mean = np}$
$=3\big(\frac{1}{3}\big)$
$=1$
$\text{Variance = npq}$
$=3\times\big(\frac{1}{3}\big)\big(\frac{2}{3}\big)$
$=\frac{2}{3}$
$\text{Mean}=1$
$\text{Variance}=\frac{2}{3}$
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Question 213 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 223 Marks
Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?
Answer
Let X be the number of people that are right-handed in the sample of 10 people.
X follows a binomial distribution with n = 10,
p = 90% = 0.9 and q = 1 - p = 0.1
$\text{P}(\text{X = r})=\text{ }^{10}\text{C}_{\text{r}}(0.9)^{\text{r}}(0.1)^{10-\text{r}}$
Probability that at most 6 are right-handed $=\text{P}(\text{X}\leq6)$
$=\text{P}(\text{X}=0)+\text{P}(\text{X}=1)+\text{P}(\text{X}=2)$
$+\text{P}(\text{X}=3)+\text{P}(\text{X}=4)+\text{P}(\text{X}=5)+\text{P}(\text{X}=6)$
$=1-\big\{\text{P}(\text{X}=7)+\text{P}(\text{X}=8)+\text{P}(\text{X}=9)+\text{P}(\text{X}=10)\big\}$
$=1-\sum\limits_{\text{r}=7}^{10}\text{ }^{10}\text{C}_\text{r}(0.9)^{\text{r}}(0.1)^{10-\text{r}}$
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Question 233 Marks
The mathematics department has 8 graduate assistants who are assigned to the same office. Each assistant is just as likely to study at home as in office. How many desks must there be in the office so that each assistant has a desk at least 90% of the time?
Answer
Let K be the number of decks and X be the number of gradute assistans in the office.
therefore, $\text{X}=8,\text{p}=\frac{1}{2},\text{q}=\frac{1}{2}$
According to the given condition,
$\text{P(X}\leq\text{k})>90\%$
$\Rightarrow\text{P(X}\leq\text{k})>0.90$
$\Rightarrow\text{P(X}>\text{k})<0.10$
$\Rightarrow\text{P(X = k}+1,\text{k}+2,\dots8)<0.10$
Therefore, $\text{P(X > 6) = P(X = 7 or X = 8})$
$\text{ }^8\text{C}_7\big(\frac{1}{2}\big)^8+\text{ }^8\text{C}_8\big(\frac{1}{2}\big)^8=0.04$
Now, $\text{P(X > 5) = P(X = 6, X = 7 or X = 8)}=0.15$
$\text{P(X > 6})<0.10$
So, if there are 6 deske then there is at least 90% chance for every graduate to get a desk.
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Question 243 Marks
If the mean and variance of a binomial variate X are 2 and 1 respectively, find P (X > 1).
Answer
$\text{Mean}=2,\text{variance}=1$
$\therefore\text{q}=\frac{\text{Variance}}{\text{Mean}}=\frac{1}{2}$
and $\text{p}=1-\frac{1}{2}=\frac{1}{2}$
$\text{n}=\frac{\text{Mean}}{\text{p}}=\frac{2}{\frac{1}{2}}=4$
The binomial distribution is given by
$\text{P(X = r})=\text{ }^4\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{4-\text{r}}$
$\therefore\text{P(X}=0)=\text{ }^4\text{C}_0\big(\frac{1}{2}\big)^0\big(\frac{1}{2}\big)^{4-0},\text{r}=0,1,2,3,4$
$=\big(\frac{1}{2}\big)^4$
$\text{P(X}>1)=1-\text{P(X}=0)$
$=1-\big(\frac{1}{2}\big)^4$
$=\frac{15}{16}$
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Question 253 Marks
An urn contains 3 white and 6 red balls. Four balls are drawn one by one with replacement from the urn. Find the probability distribution of the number of red balls drawn. Also find mean and variance of the distribution.
Answer
Let X denote the total number of red balls when four balls are drawn one by one with replacement. P(getting a red ball in one draw) $=\frac{2}{3}$ P(getting a white ball in one draw) $=\frac{1}{3}$
$\text{x}$ $0$ $1$ $2$ $3$ $4$
$\text{P(X)}$ $\big(\frac{1}{3}\big)^4$ $\frac{2}{3}\big(\frac{1}{3}\big)^3.\text{ }^4\text{C}_1$ $\big(\frac{2}{3}\big)^2\big(\frac{1}{3}\big)^2.\text{ }^4\text{C}_2$ $\big(\frac{2}{3}\big)^3\big(\frac{1}{3}\big).\text{ }^4\text{C}_3$ $\big(\frac{2}{3}\big)^4$
  $\frac{1}{81}$ $\frac{8}{81}$ $\frac{24}{81}$ $\frac{32}{81}$ $\frac{16}{81}$
Using the formula for mean, we have $\overline{\text{X}}=\sum\text{P}_{\text{i}}\text{X}_{\text{i}}$ $\text{Mean }(\overline{\text{X}})=\big(0\times\frac{1}{81}\big)+1\big(\frac{8}{81}\big)+2\big(\frac{24}{81}\big)+3\big(\frac{32}{81}\big)+4\big(\frac{16}{81}\big)$ $=\frac{1}{81}(8+48+96+64)$ $=\frac{216}{81}$ $=\frac{8}{3}$ Using the formula for variance, we have $\text{Var(X)}=\sum \text{P}_{\text{i}}\text{X}_{\text{i}}^2-\big(\sum\text{P}_{\text{i}}\text{X}_{\text{i}}\big)^2$ $\text{Var (X)}=\big\{\big(0\times\frac{1}{81}\big)+1\big(\frac{8}{81}\big)+4\big(\frac{24}{81}\big)+9\big(\frac{32}{81}\big)+16\big(\frac{16}{81}\big)\big\}-\big(\frac{8}{3}\big)^2$ $=\frac{648}{81}-\frac{64}{9}$ $=\frac{8}{9}$ Hence, the mean of the distribution is $\frac{8}{3}$ and the variance of the distribution is $\frac{8}{9}.$
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Question 263 Marks
Find the probability distribution of the number of doublets in 4 throws of a pair of dice.
Also, Find the mean and variance of this distribution.
Answer
Let X be the number of doublets in 4 throws of a pair of dice.
X follows a binomial distribution with n = 4,
$\text{p}=\text{No of getting}(1, 1)(2, 2)\dots(6,6)=\frac{6}{36}=\frac{1}{6}$
$\text{q}=1-\text{p}=\frac{5}{6}$
$\text{P}(\text{X = r})=\text{ }^4\text{C}_{\text{r}}\big(\frac{1}{6}\big)^{\text{r}}\big(\frac{5}{6}\big)^{4-\text{r}},\text{r}=0,1,2,3,4$
$\text{P}(\text{X}=0)=\text{ }^4\text{C}_0\big(\frac{1}{6}\big)^0\big(\frac{5}{6}\big)^{4-0}$
$\text{P}(\text{X}=1)=\text{ }^4\text{C}_1\big(\frac{1}{6}\big)^1\big(\frac{5}{6}\big)^{4-1}$
$\text{P}(\text{X}=2)=\text{ }^4\text{C}_2\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^{4-2}$
$\text{P}(\text{X}=3)=\text{ }^4\text{C}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^{4-3}$
$\text{P}(\text{X}=4)=\text{ }^4\text{C}_4\big(\frac{1}{6}\big)^4\big(\frac{5}{6}\big)^{4-4}$
The distribution is as follows
$\text{X}$ $0$ $1$ $2$ $3$ $4$
$\text{P(X)}$ $\frac{625}{1296}$ $\frac{500}{1296}$ $\frac{150}{1296}$ $\frac{20}{1296}$ $\frac{1}{1296}$
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Question 273 Marks
If getting 5 or 6 in a throw of an unbiased die is a success and the random variable X denotes the number of successes in six throws of the die, find $\text{P}(\text{X}\geq4).$
Answer
Let X denote the number of successes, i.e. of getting 5 or 6 in a throw of die in 6 throws.
Then, X follows a binomial distribution with n = 6;
P = of getting 5 or 6 $=\frac{1}{6}+\frac{1}{6}=\frac{1}{3};\text{q}=1-\text{p}=\frac{2}{3};$
$\text{P}(\text{X = r})=\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{6-\text{r}}$
$\text{P}(\text{X}\geq4)=\text{P}(\text{X}=4)+\text{P}(\text{X}=5)+\text{P}(\text{X}=6)$
$=\text{ }^6\text{C}_4\big(\frac{1}{3}\big)^4\big(\frac{2}{3}\big)^{6-4}+\text{ }^6\text{C}_5\big(\frac{1}{5}\big)^5\big(\frac{2}{3}\big)^{6-5}+\text{ }^6\text{C}_6\big(\frac{1}{3}\big)^6\big(\frac{2}{3}\big)^{6-6}$
$=\frac{1}{3^6}(60+12+1)$
$=\frac{73}{729}$
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Question 283 Marks
A coin is tossed 5 times. What is the probability that head appears an even number of times?
Answer
Let X be the number of heads that apper when a coin is tossed 5 times.
X follow a binomial distribution with $\text{n}=5$ and $\text{p = q}=\frac{1}{2}$
$\text{P}(\text{X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{5-\text{r}}$
$=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^5$
P (head apperars an even number of times) = P(X = 0) + P(X = 2) + P(X = 4)
$=\text{ }^5\text{C}_0\big(\frac{1}{5}\big)^5+\text{ }^5\text{C}_2\big(\frac{1}{2}\big)^5+\text{ }^5\text{C}_4\big(\frac{1}{2}\big)^5$
$=\frac{1+10+5}{2^5}$
$=\frac{16}{32}$
$=\frac{1}{2}$
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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
How many times must a man toss a fair coin so that the probability of having at least one head is more than 80%?
Answer
Let X be the number of heads and n be the minimum number of times that a man must toss a fair coin so that probability of $\text{X}\geq1$ is more than 80% and X follows a binomial distribution with
$\text{p}=\frac{1}{2},\text{q}=\frac{1}{2}$
$\text{P(X = r})=\text{ }^{\text{n}}\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{n}}$
We have $\text{P(X}\geq1)=1-\text{P(X}=0)=1-\text{ }^{\text{n}}\text{C}_0\big(\frac{1}{2}\big)^{\text{n}}=1-\frac{1}{2^{\text{n}}}$
and $\text{P(X}\geq1)>80\%$
$1-\frac{1}{2^{\text{n}}}>80\%=0.80$
$\frac{1}{2^{\text{n}}}<1-0.80=0.20$
$2^{\text{n}}>\frac{1}{0.2}=5;$
We know, $2^2<5$ while $2^3>5$
So, n = 3
So, n should be atleast 3.
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Question 313 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 323 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 333 Marks
A bag contains 7 green, 4 white and 5 red balls. If four balls are drawn one by one with replacement, what is the probability that one is red?
Answer
Let X denote the number of red balls drawn form 16 balls with replacement.
X follws a binimial distribution with $\text{n}=4,\text{p}=\frac{5}{16},\text{q}=1-\text{p}=\frac{11}{16}$
$\text{P}(\text{X = r})=\text{ }^4\text{C}_{\text{r}}\big(\frac{5}{16}\big)^{\text{r}}\big(\frac{11}{16}\big)^{4-\text{r}}$
$\text{P(one ball is red)}=\text{P}(\text{X}=1)$
$=\text{ }^4\text{C}_1\big(\frac{5}{16}\big)^1\big(\frac{11}{16}\big)^{4-1}$
$=4\big(\frac{5}{16}\big)\big(\frac{11}{16}\big)^3$
$=\frac{5}{4}\big(\frac{11}{16}\big)^3$
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Question 343 Marks
If a random variable X follows a binomial distribution with mean 3 and variance 3/2, find P (X ≤ 5).
Answer
$\text{Mean(np)}=3$ and $\text{Variance(npq)}=\frac{3}2{}$
$\therefore\text{q}=\frac{1}{2}$
and $\text{p}=1-\frac{1}{2}$
$\text{n}=\frac{\text{Mean}}{\text{p}}$
$\Rightarrow\text{n}=6$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^{6}\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{6-\text{r}},\text{r}=0,1,2\dots6$
$=\frac{\text{ }^6\text{C}_{\text{r}}}{2^6}$
$\therefore\text{P(X}\leq5)=1-\text{P(X}=6)$
$=1-\frac{1}{64}$
$=\frac{63}{64}$
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Question 353 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 $\text{r}=0,1,2\dots6$
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Question 363 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 373 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 $\text{n}=2,\text{p}=\frac{1}{2}.$
$\therefore\text{Variance = npq}=2\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{2}$
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Question 383 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 $=\text{np}$
$=10000\times\frac{1}{50}$
$=200$
$\text{Standard deviation}=\sqrt{\text{npq}}$
$=\sqrt{10000\times\frac{1}{50}\times\frac{49}{50}}$
$=14$
$\text{Expected No. of defective items}=200$
$\text{Standard deviation}=14$
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Question 393 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 $=\text{P}(\text{X}=2)$
$=\text{ }^8\text{C}_2\big(\frac{1}{10}\big)^2\big(\frac{9}{10}\big)^{8-2}$
$=\frac{28\text{ x }9^6}{10^8}$
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Question 403 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 $\text{p}=\frac{1}{5}$
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Question 413 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.
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Question 423 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 $\text{q}=1-\text{p}=\frac{1}{5}$
$\therefore\text{P(X = r})=\text{ }^{5}\text{C}_{\text{r}}\big(\frac{4}{5}\big)^{\text{r}}\big(\frac{1}{5}\big)^{5-\text{r}},\text{r}=0,1,2,\dots5$
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