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Binomial Distribution · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 32 questions.

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Question 14 Marks
An experiment succeeds twice as often as it fails. Find the probability that in the next 6 trials there will be at least 4 successes.
Answer
Let X denote the number of successes in 6 trials.
It is given that successes are twice the failures.
$\Rightarrow\text{p}=2\text{q}$
$\text{p + q}=1$
$\Rightarrow3\text{q}=1$
$\Rightarrow\text{q}=\frac{1}{3}$
$\therefore\text{p}=1-\frac{1}{3}=\frac{2}{3}$
$\text{ n}=6$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^6\text{C}_{\text{r}}\big(\frac{2}{3}\big)^{\text{r}}\big(\frac{1}{3}\big)^{6-\text{r}},\text{r}=0,1,2,\dots6$
$\text{P(atleast 4 successes})=\text{P(X}\geq4)$
$=\text{P(X}=4)+\text{P(X}=5)+\text{P(X}=6)$
$\text{ }^6\text{C}_4\big(\frac{2}{3}\big)^4\big(\frac{1}{3}\big)^{6-4}+\text{ }^6\text{C}_5\big(\frac{2}{3}\big)^5\big(\frac{1}{3})^{6-5}+\text{ }^6\text{C}_6\big(\frac{2}{3}\big)^6\big(\frac{1}{3}\big)^{6-6}$
$=\frac{15(2^4)+6(32)+64}{3^6}$
$=\frac{240+192+64}{729}$
$=\frac{496}{729}$
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Question 24 Marks
If the mean and variance of a random variable X having  a binomial distribution are 4 and 2 respectively, find P (X = 1).
Answer
Let n and p be the parmeters of binomial distribution,
Given,
$\text{Mean = np}=4$
$\text{Variance = npq}=2$
Dividing equation (2) by (1),
$\frac{\text{npq}}{\text{np}}=\frac{2}{4}$
$\text{q}=\frac{1}{2}$.
$\text{p}=1-\frac{1}{2}$
$\text{p}=\frac{1}{2}$
Put the value of p in equation (1),
$\text{np}=4$
$\text{n}\big(\frac{1}{2}\big)=4$
$\text{n}=8$
Hence, binomial distribution is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X= r})\text{ }^8\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{8-\text{r}}$
$\text{P(X=}1)$
$=\text{ }^8\text{c}_{1}\big(\frac{1}{2}\big)^{1}\big(\frac{1}{2}\big)^{8-1}$
$=8\big(\frac{1}{2}\big)^8$
$=\big(\frac{1}{2}\big)^5$
$=\frac{1}{32}$
$\text{P(X}=1)=\frac{1}{32}$
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Question 34 Marks
A bag contains 10 balls, each marked with one of the digits from 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?
Answer
Let p denote the probability of getting a ball market with 0. So
$\text{p}=\frac{1}{10}$ [Since balls are market with 0, 1, 2, 3, 4, 5, 6, 7, 8, 8, 9]
$\text{q}=1-\frac{1}{10}$ [Since p + q = 1]
$\text{q}=\frac{9}{10}$
Let X denote the variable presenting the number of balls marked with 0 out of four balls drawn. probability of drawing r balls out of n balls that are marked 0 is given by
$\text{P}(\text{X = r})=\text{ }^\text{n}\text{c}_\text{r}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^{4}\text{c}_\text{r}\big(\frac{1}{10}\big)^\text{r}\big(\frac{9}{10}\big)^{4-\text{r}}\dots(1)$
Probability of getting none balls marked with 0
$=\text{P}(\text{X}=0)$
$=\text{ }^4\text{c}_0\big(\frac{1}{10}\big)^0\big(\frac{9}{10}\big)^{4-0}$
$=1.1.\big(\frac{9}{10}\big)^4$
$=\big(\frac{9}{10}\big)^4$
Probability of getting nine balls marked with $0=\big(\frac{9}{10}\big)^4$
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Question 44 Marks
In a large bulk of items, 5 percent of the items are defective. What is the probability that a sample of 10 items will include not more than one defective item?
Answer
Let X denote the number of defective items in a sample of 10 items.
X follows a binomial distribution with n = 10;
P = Probability of detective items = 5% = 0.05; q = 1- p = 0.95
P(X = r)=Cr10(0.05)r(0.95) 10 - r
P(x=r) = 10Cr(0.05)r (0.95)10-r
Probability (sample of 10 items will include not more than one defective item)=P(X ≤ 1)
= P(X = 0) + P(X = 1)
= 10C0(0.05)0 (0.95)10-0 + 10C1(0.05)1 (0.95)10-1
= (0.95)9 (0.95+0.5)
= 1.45(0.95)9
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Question 54 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.
Answer
Let p be the probability of getting a doublet in a throw of a pair of dice, so
$\text{p}=\frac{6}{36}$ [Since (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]
$=\frac{1}{6}$
$\text{q}=1-\frac{1}{6}$ [Since p + q = 1]
$=\frac{5}{6}$
Let X denote the number of grtting doublets i.e. success out of 4 times. So, probility distribution is given by
$\text{X}$
 $\text{P(X)}$
$0$
 $\text{ }^4\text{C}_0\big(\frac{1}{6}\big)^0\big(\frac{5}{6}\big)^{4-0}=\big(\frac{5}{6}\big)^4$
$1$
 $\text{ }^4\text{C}_1\big(\frac{1}{6}\big)^1\big(\frac{5}{6}\big)^{4-1}=4\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^3=\frac{2}{3}\big(\frac{5}{6}\big)^3$
$2$
 $\text{ }^4\text{C}_2\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^{4-2}=\frac{4.3}{2}\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^2=\frac{25}{216}$
$3$
 $\text{ }^4\text{C}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^{4-3}=\frac{4.3}{2}\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)=\frac{5}{324}$
$4$
 $\text{ }^4\text{C}_4\big(\frac{1}{6}\big)^4\big(\frac{5}{6}\big)^{4-4}=\big(\frac{1}{6}\big)^4=\frac{1}{1296}$
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Question 64 Marks
In a hospital, there are 20 kidney dialysis machines and that the chance of any one of them to be out of service during a day is 0.02. Determine the probability that exactly 3 machines will be out of service on the same day.
Answer
let X = nimber of out of service machines
p = probability that machine will be out of service on the
same day $=\frac{2}{100}$
q = probability that machine will be in service on the
same day $=\frac{8}{100}$
$\text{P(X}=3)=$ probability exactly 3 machines will be out of service on the same day
$\text{P(X}=3)=\text{ }^{20}\text{C}_3\times\big(\frac{2}{100}\big)^3\big(\frac{8}{100}\big)^0$
$=1140\times0.000008=0.00912$
For low probability event Poisson' distribution is used instead of Binomial distribution. then,
$\lambda=\text{np}=20\times0.02=0.4$
$\text{P(X = r})=\frac{\text{e}^{-\lambda}\times\lambda^3}{\text{r}!}$
$\text{P(X}=3)=\frac{\text{e}^{-0.4}\times0.4^3}{3!}=0.6703\times\frac{0.064}{6}=0.0071$
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Question 74 Marks
A coin is tossed 5 times. What is the probability of getting at least 3 heads?
Answer
Probability of getting head on one throw of coin $=\frac{1}{2}$
So, $\text{p}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}$
$\text{q}=\frac{1}{2}$ [Since p + q = 1]
The coin is tossed 5 times. Let x denote the number of getting head as 5 tosses of coins.
So probability of getting r head in n tosses of coin is given by
$\text{P}(\text{x = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n - r}}$
$\text{P}(\text{x = r})=\text{ }^5\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{5-\text{r}}\dots(1)$
probability of getting at least 3 heads
$=\text{P}(\text{X}=3)+\text{P}(\text{x}=4)+\text{p}(\text{x}=5) $
$=\text{ }^5\text{c}_3\big(\frac{1}{2}\big)^3.\big(\frac{1}{2}\big)^{5-3}+\text{ }^5\text{c}_4\big(\frac{1}{2}\big)^4\big(\frac{1}{2}\big)^{5-4}+\text{ }^5\text{c}_5\big(\frac{1}{2}\big)^5\big(\frac{1}{2}\big)^0$ [Using (1)]
$=\text{ }^5\text{c}_3\big(\frac{1}{2}\big)^3\big(\frac{1}{2}\big)^2+\text{ }^5\text{c}_4\big(\frac{1}{2}\big)^4\big(\frac{1}{2}\big)+\text{ }^5\text{c}_5\big(\frac{1}{2}\big)^5.1$
$=\frac{5.4}{2}.\big(\frac{1}{2}\big)^5+5\big(\frac{1}{2}\big)^5+1.\big(\frac{1}{2}\big)^5$
$=\big(\frac{1}{2}\big)^5[10+5+1]$
$=16.\frac{1}{32}$
$=\frac{1}{2}$
The required probability is $=\frac{1}{2}$
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Question 84 Marks
Find the probability of 4 turning up at least once in two tosses of a fair die.
Answer
Let p denote the 4 turning up in a toss of a fair die, So
$\text{p}=\frac{1}{6}$
$\text{q}=1-\frac{1}{6}$
$\text{q}=\frac{5}{6}$ [Since p + q = 1]
Let X denote the variable showing the number of turning 4 up in 2 tosses of die.
Probability of getting 4, r times in n tosses of a die is given by
$\text{p}(\text{X = r})=\text{ }^{\text{n}}\text{c}_\text{r}\text{p}^\text{r}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^2\text{c}_\text{r}\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^{2-\text{r}}\dots(1)$
Probability of getting 4 at least once in tow tosses of a fair die
$=\text{P}(\text{X}=1)+\text{P}(\text{X}=2)$
$=1-\text{P}(\text{X}=0)$
$=1-\Big[\text{ }^2\text{c}_0\big(\frac{1}{6}\big)^0\big(\frac{5}{6}\big)^{2-0}\Big]$ [Using (1)]
$=1-\Big[1.1.\big(\frac{5}{6}\big)^2\Big]$
$=1-\Big[\frac{25}{36}\Big]$
$=\frac{36-25}{36}$
$=\frac{11}{36}$
So,
Required probability $=\frac{11}{36}$
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Question 94 Marks
cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the mean and variance of red cards.
Answer
It is given that the cards are drawn successively with replacement so the events are indepent.

Therefore, the drawing of the cards follow binomial distribution.

Probability of drawng a red card $=\text{p}=\frac{26}{52}=\frac{1}{2}$

$\therefore\text{q}=1-\text{p}=1-\frac{1}{2}=-\frac{1}{2}$

Also, $\text{n}=3$

Let X be the random variable denoting the number of red cards drawn from a well shuffied pack of 52 cards.

$\therefore\text{P(X = r})=\text{ }^3\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{3-\text{r}}\big(\frac{1}{2}\big)^{\text{r}}=\text{ }^3\text{C}_{\text{r}}\big(\frac{1}{2}\big)^3,\text{r}=0,1,2,3$

Probability of drawing no red ball $=\text{P(X}=0)=\text{ }^3\text{C}_0\big(\frac{1}{2}\big)^3=\frac{1}{8}$

Probability of drawing one red ball $=\text{P(X}=1)=\text{ }^3\text{C}_1\big(\frac{1}{2}\big)^3=\frac{3}{8}$

Probability of drawing tow red balls $=\text{P(X}=2)=\text{ }^3\text{C}_2\big(\frac{1}{2}\big)^3=\frac{3}{8}$

Probability of drawing three red balls $=\text{P(X}=3)=\text{ }^3\text{C}_3\big(\frac{1}{2}\big)^3=\frac{1}{8}$

Thus, the probability distribution of X is as follows:

$\text{x}_{\text{i}}$

 $\text{p}_{\text{i}}$

 $\text{p}_{\text{i}}\text{x}_{\text{i}}$

 $\text{p}_{\text{i}}\text{x}_{\text{i}}^2$
$0$

 $\frac{1}{8}$

 $0$

 $0$
$1$

 $\frac{3}{8}$

 $\frac{3}{8}$

 $\frac{3}{8}$
$2$

 $\frac{3}{8}$

 $\frac{6}{8}$

 $\frac{12}{8}$
$3$

 $\frac{1}{8}$

 $\frac{3}{8}$

 $\frac{9}{8}$

 

 

 $\sum\text{p}_{\text{i}}\text{x}_{\text{i}}=\frac{12}{8}$

 $\sum\text{p}_{\text{i}}\text{x}_{\text{i}}^2=3$
Mean of X $=\sum\text{p}_{\text{i}}\text{x}_{\text{i}}=\frac{12}{8}=\frac{3}{2}$

variance of X $=\sum\text{p}_{\text{i}}\text{x}_{\text{i}}^2-(\text{Mean})^2=3-\frac{9}{4}=\frac{3}{4}$

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Question 104 Marks
The probability that a student entering a university will graduate is 0.4. Find the probability that out of 3 students of the university.
  1. none will graduate.
  2. only one will graduate.
  3. all will graduate.
Answer
Let X be the number of students that gradute from among 3 students.
Let p = probability that a student entering a university will garduate.
Here, n = 3, p = 0.4 and q = 0.6
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^3\text{C}_{\text{r}}(0.4)^{\text{r}}(0.6)^{3-\text{r}},\text{r}=0,1,2,3$
  1. $\text{P(X}=0)=\text{q}^3=0.216$
  2. $\text{P(X}=1)=3(0.4)(0.36)=0.432$
  3. $\text{P(X}=3)=\text{p}^3=0.064$
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Question 114 Marks
The probability of a shooter hitting a target is $\frac{3}{4}.$ How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?
Answer
Let the shooter fire n times.
n fires are bernoulli trials.
In each trial, p = probability of hitting the target $=\frac{3}{4}$
And q = probability of not hitting the target $=1-\frac{3}{4}=\frac{1}{4}$
Then, $\text{P(X = x})=\text{ }^{\text{n}}\text{c}_{\text{x}}\text{q}^{\text{n}-\text{x}}\text{p}^{\text{x}}=\text{ }^{\text{n}}\text{c}_{\text{x}}\big(\frac{1}{4}\big)^{\text{n}-\text{x}}\big(\frac{3}{4}\big)^{\text{x}}=\text{ }^{\text{n}}\text{c}_{\text{x}}\frac{3^{\text{x}}}{4^{\text{n}}}$
Now, given that
P(hitting the target atleast once) > 0.99
i.e. $\text{P(X}\geq1)>0.99$
$\Rightarrow1-\text{P(X}=0)>0.99$
$\Rightarrow1-\text{ }^{\text{n}}\text{c}_0\frac{1}{4^{\text{n}}}>0.99$
$\Rightarrow\text{ }^{\text{n}}\text{c}_0\frac{1}{4^{\text{n}}}<0.01$
$\Rightarrow\frac{1}{4^{\text{n}}}<0.01$
$\Rightarrow4^{\text{n}}>\frac{1}{0.01}=100$
The minimum value of n to satisfy this inequality is 4
Thus, the shooter must fire 4 times.
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Question 124 Marks
A box contains 100 tickets, each bearing one of the numbers from 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear numbers divisible by 10.
Answer
Let P denote the probability of getting a ticket bearing number divisibal by 10, So
$\text{p}=\frac{10}{100}$ [Since there are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 which are divisible by 10] 
$\text{p}=\frac{1}{10}$
$\text{q}=1-\frac{1}{10}$ [Since p + q = 1]
$\text{q}=\frac{9}{10}$
Let X denote the variable representing the number of tickets bearing a number divisible by 10 out of 5 tickets. probability of getting r tickets bearing a number divisible by 10 out ot n tickets is given by
$\text{P}(\text{X = r})=\text{ }^\text{n}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^{\text{5}}\text{c}_{\text{r}}\big(\frac{1}{10}\big)^{\text{r}}\big(\frac{9}{10}\big)^{5-\text{r}}\dots(1)$
Probability of getting all the tickets bearing a number divisible by 10
$=\text{ }^5\text{c}_5\big(\frac{1}{10}\big)^5\big(\frac{9}{10}\big)^{5-5}$ [Using (1)]
$=1.\big(\frac{1}{10}\big)^5\big(\frac{9}{10}\big)^0$
$=\big(\frac{1}{10}\big)^5$
Required probability $=\big(\frac{1}{10}\big)^5$
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Question 134 Marks
Five bad oranges are accidently mixed with 20 good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence find the mean and variance of the distribution.
Answer
Let X be the random variable denoting the number of bad oranges drawn.
P (getting a good orange) $=\frac{20}{25}=\frac{4}{5}$
P (getting a bad orange) $=\frac{5}{25}=\frac{1}{5}$
The probability distribution of X is given by
$\text{X}$
 $0$
 $1$
 $2$
 $3$
 $4$
$\text{P(X)}$
 $\big(\frac{4}{5}\big)^4=\frac{256}{625}$
 $\text{ }^4\text{C}_1\big(\frac{4}{3}\big)^3\big(\frac{1}{5}\big)=\frac{256}{625}$
 $\text{ }^4\text{C}_2\big(\frac{4}{5}\big)^2\big(\frac{1}{5}\big)^2=\frac{96}{625}$
 $\text{ }^4\text{C}_3\big(\frac{4}{5}\big)\big(\frac{1}{5}\big)^3=\frac{16}{625}$
 $\big(\frac{1}{5}\big)^4=\frac{1}{625}$
Mean of X is given by
$\overline{\text{X}}=\sum\text{P}_{\text{i}}\text{X}_{\text{i}}$
$=0\times\frac{256}{625}+1\times\frac{256}{625}+2\times\frac{96}{625}+3\times\frac{16}{625}+4\times\frac{1}{625}$
$=\frac{1}{625}(256+192+48+4)$
$=\frac{4}{5}$
Variance of X given by
$\text{Var (X)}=\sum\text{P}_{\text{i}}\text{X}_{\text{i}}^2-\big(\sum\text{P}_{\text{i}}\text{X}_{\text{i}}\big)^2$
$=0\times\frac{256}{625}+1\times\frac{256}{625}+4\times\frac{96}{625}+9\times\frac{16}{625}+16\times\frac{1}{625}-\big(\frac{4}{5}\big)^2$
$=\frac{1}{625}(256+384+144+16)-\frac{16}{25}$
$=\frac{800}{625}-\frac{16}{25}$
$=\frac{400}{625}$
$=\frac{16}{25}$
Thus, the mean and vairance of the distribution are $\frac{4}{5}$ and $\frac{16}{25},$ respectively.
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Question 144 Marks
Find the probability distribution of the number of doublets in three throws of a pair of dice and find its mean.
Answer
Throwing a doublet i.e. $\big\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\big\}$
Total number of outcomes = 36
Let p be the probability of success therefore
$\text{p}=\frac{6}{3}6=\frac{1}{6}$
Let q be the probability of failure therefore $\text{q}=1-\text{p}=1-\frac{1}{6}=\frac{5}{6}$
Since there is three rows of dise so n = 3
Let X be the random variable for getting doublet, therefore X can take at max 3 values.
$\text{P(X}=0)=\text{ }^3\text{c}_0\text{p}^0\text{q}^{3}=\big(\frac{5}{6}\big)^{3}=\frac{125}{216}$
$\text{P(X}=1)=\text{ }^3\text{c}_1\text{p}^1\text{q}^2=3.\frac{1}{6}.\big(\frac{5}{6}\big)^2=\frac{75}{216}$
$\text{P(X}=2)=\text{ }^3\text{c}_2\text{p}^2\text{q}^1=3.\big(\frac{1}{6}\big)^2.\big(\frac{5}{6}\big)=\frac{15}{216}$
$\text{P(X}=3)=\text{ }^3\text{c}_3\text{p}^3\text{q}^0=\big(\frac{1}{6}\big)^3=\frac{1}{216}$
Mean
$\mu=\sum^3_{\text{i}-1}\text{X}_{\text{i}}\text{P(X}_{\text{i}})=0.\frac{125}{216}+1.\frac{75}{216}+2.\frac{15}{216}+3.\frac{1}{216}$
$=\frac{75+30+3}{216}=\frac{108}{216}=\frac{1}{2}$
Hence the mean is $=\frac{1}{2}$
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Question 154 Marks
In a binomial distribution the sum and product of the mean and the variance are $\frac{25}{3}$ and $\frac{50}{3}$ respectively. Find the distribution.
Answer
Let n and p be the parameter of distribution binomial distribution. So
$\text{q}=1-\text{p}$ as $\text{p + q}=1$
$\text{ Mean + variance}=\frac{25}{3}$
$\text{np + npq}=\frac{25}{3}$
$\text{np(1+ q)}=\frac{25}{3}$
$\text{np}=\frac{25}{3(1+\text{ q})}\dots(1)$
$\text{Mean}\times\text{Variance}=\frac{50}{3}$
$\text{np}\times\text{npq}=\frac{50}{3}$
$\text{n}^2\text{p}^2\text{q}=-\frac{50}{3}$
$\Big[\frac{25}{3(1+\text{q})}\Big].\text{q}=\frac{50}{3}$ [using(1)]
$625\text{q}=\frac{50}{3}\big[9(1+\text{q})^2\big]$
$625\text{q}=150(1+\text{q})^2$
$25\text{q}=6(1+\text{q})^2$
$6+6\text{q}^2+12\text{q}-25\text{q}=0$
$6\text{q}^2-13\text{q}+6=0$
$6\text{q}^2-9\text{q}-4\text{q}+6=0$
$3\text{q}(2\text{q}-3)-2(2\text{q}-3)=0$
$(2\text{q}-3)(3\text{q}-2)=0$
$\Rightarrow2\text{q}-3=0$ or $3\text{q}-2=0$
$\Rightarrow\text{q}=\frac{3}{2}$ or $\text{q}=\frac{2}{3}$
Since $\text{q}\leq1,$ So
$\text{q}=\frac{2}{3}$
$\text{p}=1-\text{q}$
$=1-\frac{2}{3}$
$\text{p}=\frac{1}{3}$
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Question 164 Marks
Find the binomial distribution whose mean is 5 and variance $\frac{10}{3}.$
Answer
Let n and p be the parameters of binomial distribution.
Given, $\text{Mean = np}=5\dots(1)$
$\text{Variance = npq}=\frac{10}{3}\dots(2)$
Dividing (2) by (1)
$\frac{\text{npq}}{\text{np}}=\frac{\frac{10}{3}}{5}$
$\text{q}=\frac{2}{3}$
So, $\text{p}=1-\text{q}$ [Since p + q = 1]
$=1-\frac{2}{3}$
$\text{p}=\frac{1}{3}$
Put the value of p in equation (1),
$\text{np}=5$
$\text{n}=5\times3$
$\text{n}=15$
Hence, the binomial distribution is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X = r})=\text{ }^{15}\text{c}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{15-\text{r}}$
$\text{r}=0,1,2,\dots15$
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Question 174 Marks
Determine the binomial distribution whose mean is 9 and variance $\frac{9}{4}.$
Answer
Let X denote the variance with parameters n and p

$\text{p + q}=1$

$\text{q}=1-\text{p}$

Given,

$\text{Mean = np} =9\dots(1)$

$\text{variance = npq}=\frac{9}{4}\dots(2)$

$\frac{\text{npq}}{\text{np}}=\frac{\frac{9}{4}}{9}$ [By diving (1) by (2)]

$\text{q}=\frac{1}{4}$

So, $\text{p}=1-\text{q}$

$=1-\frac{1}{4}$

$\text{p}=\frac{3}{4}$

Put p in equation (1),

$\text{n}\big(\frac{3}{4}\big)=9$

$\Rightarrow\text{n}=\frac{36}{3}$

So, $\text{n}=12$

The distribution is given by

$=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}(\text{q})^{\text{n}-\text{r}}$

$\text{P(X = r})=\text{ }^{12}\text{c}_{\text{r}}\big(\frac{3}{4}\big)^{\text{r}}\big(\frac{1}{4}\big)^{12-\text{r}}$

$\text{for r}=0,1,2,\dots12$

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Question 184 Marks
An unbiased coin is tossed 8 times. Find, by using binomial distribution, the probability of getting at least 6 heads.
Answer
Binomial Distribution formula is given by
$\text{P(X)}=\text{ }^{\text{n}}\text{c}_{\text{x}}\text{p}^{\text{x}}\text{q}^{\text{n}-\text{x}},$ where $\text{X}=0,1,2,\dots\text{n}$
Let X = No. of heads in a toss
We need probability of 6 or more heads
$\text{X}=6,7,8$
Here $\text{p}=\frac{1}{2}$ and $\text{q}=\frac{1}{2}$
P(6)=Prob of getting 6 heads, 2 tails $=\text{ }^8\text{C}_6\big(\frac{1}{6}\big)^6\times\big(\frac{1}{2}\big)^2$
P(7) = Prob of getting 7 heads, 1 tails $=\text{ }^8\text{C}_7\big(\frac{1}{2}\big)^7\times\big(\frac{1}{2}\big)^1$
P(8) = Prob of getting 8 heads, 0 tails $=\text{ }^8\text{C}_8\big(\frac{1}{2}\big)^8\times\big(\frac{1}{2}\big)^0$
The probability of getting at least 6 heads (not more than 2 tails) is then
$\text{ }^8\text{C}_6\big(\frac{1}{6}\big)^6\times\big(\frac{1}{2}\big)^2+\text{ }^8\text{C}_1\big(\frac{1}{2}\big)^7\times\big(\frac{1}{2}\big)^1+\text{ }^8\text{C}_2\big(\frac{1}{2}\big)^8\times\big(\frac{1}{2}\big)^0$
$=\frac{1}{256}+8\frac{1}{256}+28\frac{1}{256}=\frac{37}{256}$
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Question 194 Marks
Determine the binomial distribution whose mean is 20 and variance 16.
Answer
Given that,
$\text{Mean = np}=20\dots(1)$
$\text{Variance = npq}=16\dots(2)$
Let n and p be the parameters of distribution dividing equation (2) by (1)
$\frac{\text{npq}}{\text{np}}=\frac{16}{20}$
$\text{q}=\frac{4}{5}$
So, $\text{p}=1-\text{q}$ [Since p + q = 1]
$=1-\frac{4}{5}$
$\text{p}=\frac{1}{5}$
Put p in equation (1),
$\text{np}=20$
$\text{n}\big(\frac{1}{5}\big)=20$
$\text{n}=20\times5$
$\text{n}=100$
So, binomial distribution is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X = r})=\text{ }^{100}\text{c}_{\text{r}}\big(\frac{1}{5}\big)^{\text{r}}\big(\frac{4}{5}\big)^{100-\text{r}}$
$\text{r}=0,1,2,3,\dots100$
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Question 204 Marks
The probability that a certain kind of component will survive a given shock test is $\frac{3}{4}.$ Find the probability that among 5 components tested.
  1. exactly 2 will survive.
  2. at most 3 will survive.
Answer
Let p be the probability that componet survive the shock test.So

$\text{p}=\frac{3}{4}$

$\text{q}=1-\frac{3}{4}$ [Since p + q = 1]

$\text{q}=\frac{1}{4}$

Let X denote the random variable representing the number of components that survive out of n components is given by

$\text{P(X = r } ) \ =\text{ }^{\text{n}}\text{c}_{\text{r}}\big(\frac{3}{4}\big)^{\text{r}}\big(\frac{1}{4}\big)^{5-\text{r}}\dots(1)$

  1. Probability that exactly 2 will survive the shock test

$=\text{P(X}=2)$

$=\text{ }^5\text{C}_2\big(\frac{3}{4}\big)^2\big(\frac{1}{4}\big)^{5-2}$

$=\frac{5.4}{2}\big(\frac{9}{16}\big)\big(\frac{1}{64}\big)$

$=\frac{45}{512}=0.0879$

Probability that exactly 2 survive $=0.0879$

  1. P( atmost 3 will survive) $=\text{P(X}\leq3)$

$=\text{P(X}=0)+\text{P(X}=1)+\text{P(X}=2)+\text{P(X}=3)$

$=\text{ }^5\text{C}_0\big(\frac{3}{4}\big)^0\big(\frac{1}{4}\big)^{5-0}+\text{ }^5\text{C}_1\big(\frac{3}{4}\big)^1\big(\frac{1}{4}\big)^{5-1}$

$+\text{ }^5\text{C}_2\big(\frac{3}{4}\big)^2\big(\frac{1}{4}\big)^{5-2}+\text{ }^5\text{C}_3\big(\frac{3}{4}\big)^3\big(\frac{1}{4}\big)^{5-3}$

$=\big(\frac{1}{4}\big)^5+5\big(\frac{3}{4}\big)\big(\frac{1}{4}\big)^4+10\big(\frac{3}{4}\big)^2\big(\frac{1}{4}\big)^3+10\big(\frac{3}{4}\big)^3\big(\frac{1}{4}\big)^2$

$=\frac{1+15+90+270}{1024}$

$=\frac{376}{1024}$

$=0.3672$

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Question 214 Marks
A card is drawn and replaced in an ordinary pack of 52 cards. How many times must a card be drawn so that.
  1. there is at least an even chance of drawing a heart.
  2. the probability of drawing a heart is greater than $\frac{3}{4}$?
Answer
Let p denote the probability of drawing a heart from a deck of 52 cards, so

$\text{p}=\frac{13}{52}$ [$\because$ There are 13 hearts in deck]

$\text{p}=\frac{1}{4}$

$\text{q}=1-\frac{1}{4}$ [since p + q = 1]

$\text{q}=\frac{3}{4}$

Let the card is drawn n times. so Binomial distribution is given by 

$\text{P(X = r)}=\text{ }^{\text{n}}\text{c}_\text{r}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$

where X denote the number of spades drawn and r = 0, 1, 2, 3, ..... n

  1. We have to find the smallest value of n for which P(X = 0) is less than $\frac{1}{4}$

$\text{P(X = 0)}<\frac{1}{4}$

$\text{ }^\text{n}\text{C}_0\big(\frac{1}{1}\big)^0\big(\frac{3}{4}\big)^{\text{n}-0}<\frac{1}{4}$

$\big(\frac{3}{4}\big)^{\text{n}}<\frac{1}{4}$

Put  $\text{n}=1, \big(\frac{3}{4}\big)\nless\frac{1}{4}$

$\text{n}=2, \big(\frac{3}{4}\big)^2\nless\frac{1}{4}$

$\text{n}=3, \big(\frac{3}{4}\big)^3\nless\frac{1}{4}$

So, smallest value of n = 3

$\therefore$ we must draw card at least 3 times

  1. Given the probability of drawing a heart $>\frac{3}{4}$

$1-\text{P(X = 0)}>\frac{3}{4}$

$1-\text{ }^{\text{n}}\text{C}_0\big(\frac{1}{4}\big)^0\big(\frac{3}{4}\big)^{\text{n}-0}>\frac{3}{4}$

$1-\big(\frac{3}{4}\big)^{\text{n}}>\frac{3}{4}$

$1-\frac{3}{4}>\big(\frac{3}{4}\big)^{\text{n}}$

$\frac{1}{4}>\big(\frac{3}{4}\big)^{\text{n}}$

For $\text{n}=1,\big(\frac{3}{4}\big)^1$ not less than $\frac{1}{4}$

$\text{n}=2,\big(\frac{3}{4}\big)^2$ not less than $\frac{1}{4}$

$\text{n}=3,\big(\frac{3}{4}\big)^3$ not less than $\frac{1}{4}$

$\text{n}=4,\big(\frac{3}{4}\big)^4$ not less than $\frac{1}{4}$

$\text{n}=5,\big(\frac{3}{4}\big)^5$ not less than $\frac{1}{4}$

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Question 224 Marks
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is $\frac{1}{100}.$ What is the probability that he will win a prize.
  1. at least once.
  2. exactly once.
  3. at least twice.
Answer
Let X denote the number of times the person wine the lottery.

Then, X follows a binomial distribution with n = 50.

Let p be the probability of winning a prize.

$\therefore\text{p}=\frac{1}{100},\text{q}=1-\frac{1}{100}=\frac{9}{100}$

Hence, the distribution is given by

$\text{P(X = r})=\text{ }^{50}\text{C}_{\text{r}}\big(\frac{1}{100}\big)^{\text{r}}\big(\frac{99}{100}\big)^{50-\text{r}},\text{r}=0,1,2,\dots50$

  1. P(winning at least once) $=\text{P(X}\geq0)$

$=1-\text{P(X}-0)$

$=1-\big(\frac{99}{100}\big)^{50}$

  1. P(winning exactly once) $=\text{P(X}=1)$

$=\text{ }^{50}\text{C}_1\big(\frac{1}{100}\big)^1\big(\frac{99}{100}\big)^{50-1}$

$=\frac{1}{2}\big(\frac{99}{100}\big)^{49}$

  1. P(winning at least twice) $=\text{P(X}\geq2)$

$=1-\text{P(X}=0)-\text{P(X}=1)$

$=1-\big(\frac{99}{100}\big)^{50}-\text{ }^{50}\text{C}_1\times\frac{1}{100}\times\big(\frac{99}{100}\big)^{49}$

$=1-\frac{99^{49}\times149}{100^{50}}$

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Question 234 Marks
The mean and variance of a binomial variate with parameters n and p are 16 and 8, respectively. Find P(X = 0), P (X = 1) and P (X ≥ 2).
Answer
Given that, parameter for binomial distribution are n and p.
Also, $\text{Mean = np}=16\dots(1)$
$\text{Variance = npq = 8}\dots(2)$
Dividing (2) ane (1)
$\text{q}=\frac{1}{2}$
So, $\text{p}=1-\frac{1}{2}$ [as p + q = 1]
$\text{p}=\frac{1}{2}$
put the value of p in equation (1),
$\text{np}=16$
$\text{n}\big(\frac{1}{2}\big)=16$
$\text{n}=32$
Hence, binomial distribution is given by,
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{32-\text{r}}\dots(3)$
$\text{P(X}=0)$
$=\text{ }^{32}\text{c}_0\big(\frac{1}{2}\big)^0\big(\frac{1}{2}\big)^{32-0}$ [Using (3)]
$=\big(\frac{1}{2}\big)^{32}$
$=\text{P(X}=1)$
$=\text{ }^{32}\text{c}_1\big(\frac{1}{2}\big)^1\big(\frac{1}{2}\big)^{32-1}$
$=32.\frac{1}{2}\big(\frac{1}{2}\big)^{31}$
$=\big(\frac{1}{2}\big)^{27}$
$\text{P(X}\geq2)$
$=1-\big[\text{P(X}=0)+\text{P(X}=1)\big]$
$=1-\Big[\big(\frac{1}{2}\big)^{32}+\big(\frac{1}{2}\big)^{27}\Big]$
$=1-\big(\frac{1}{2}\big)^{27}\big(\frac{1}{32}+1\big)$
$=1-\big(\frac{1}{2}\big)^{27}\big(\frac{33}{32}\big)$
$=1-\frac{33}{2^{32}}$
Hence
$\text{P(X}=0)=\big(\frac{1}{2}\big)^{32},\text{P(X = 1})=\big(\frac{1}{2}\big)^{27},\text{P(X}\geq2)=1-\frac{33}{2^{32}}$
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Question 244 Marks
If X follows a binomial distribution with mean 4 and variance 2, find P (X ≥ 5).
Answer
Let n and p the parameter of binomial distribution.
Given,
$\text{Mean = np}=4\dots(1)$
$\text{Variance = npq}=2\dots(2)$
Dividing equation (2) by (1),
$\frac{\text{npq}}{\text{np}}=\frac{2}{4}$
$\text{q}=\frac{1}{2}$
$\text{p}=1-\frac{1}{2}$ [Since p + q = 1]
$\text{p}=\frac{1}{2}$
Put the value of p in equation (1),
$\text{np}=4$
$\text{n}\big(\frac{1}{2}\big)=4$
$\text{n}=8$
Hence, binomial distributon is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X = r})=\text{ }^8\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{8-\text{r}}\dots(3)$
$\text{P(X}\geq5)$
$=\text{P(X}=5)+\text{P(X}=6)+\text{P(X}=7)+\text{P(X}=8)$
$=\text{ }^{8}\text{C}_5\big(\frac{1}{2}\big)^{5}\big(\frac{1}{2}\big)^{3}+\text{ }^{8}\text{C}_6\big(\frac{1}{2}\big)^{6}\big(\frac{1}{2}\big)^{2}+\text{ }^{8}\text{C}_7\big(\frac{1}{2}\big)^{7}\big(\frac{1}{2}\big)+\text{ }^{8}\text{C}_8\big(\frac{1}{2}\big)^{8}$
[Using equation (3)]
$=\frac{8\times7\times6}{3\times2}\big(\frac{1}{2}\big)^8+\frac{8\times7}{2}\big(\frac{1}{2}\big)^8+8\big(\frac{1}{2}\big)^8+\big(\frac{1}{2}\big)^8$
$=\big(\frac{1}{2}\big)^8\big[56+28+8+1\big]$
$=\frac{93}{256}$
$\text{P(X}\geq5)=\frac{93}{256}$
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Question 254 Marks
It is known that 60% of mice inoculated with a serum are protected from a certain disease. If 5 mice are inoculated, find the probability that.
  1. none contract the disease.
  2. more than 3 contract the disease.
Answer
Let X be the number of mice that contract the disease.

Then, X follows a binomial distribution with n = 5.

Let p be the probability of mice that cantract the disease.

$\therefore\text{p}=0.4$ and $\text{q}=0.6$

Hence, the distribution is given by

$\text{P(X = r})=\text{ }^5\text{C}_{\text{r}}(0.4)^{\text{r}}(0.6)^{5-\text{r}},\text{r}=0,1,2,3,4,5$

  1. $\text{P(X}=0)=\text{ }^5\text{C}_0(0.4)^0(0.6)^{5-0}$

$=(0.6)^5$

$=0.0778$

  1.  $\text{P(X}>3)=\text{P(X}=4)+\text{P(X}=5)$

$=\text{ }^5\text{C}_4(0.4)^4(0.6)^{5-4}+\text{ }^5\text{C}_5(0.4)^5(0.6)^{5-5}$

$=0.0768+0.01024$

$=0.08704$

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Question 264 Marks
From a lot of 15 bulbs which include 5 defective, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.
Answer
Let getting a defective bulb in a trial be a success.
We have,
p = probability of getting a defective bulb $=\frac{5}{15}=\frac{1}{3}$ and
q = probability of getting non - defective bulb $=1-\text{p}=1-\frac{1}{3}=\frac{2}{3}$
Let X denote the number of success in a sample of 4 trials. Then,
X follows binomial distribution with parameter $\text{n}=4$ and $\text{p}=\frac{1}{3}$
$\therefore\text{P(X = r})=\text{ }^4\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(4-\text{r})}=\text{ }^4\text{c}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{(4-\text{r})}=\frac{\text{ }^4\text{c}_{\text{r}}2^{(4-\text{r})}}{3^4},$ where $\text{r}=0,1,2,3,4$
i.e.
$\text{P(X}=0)=\frac{\text{ }^4\text{C}_02^4}{3^4}=\frac{16}{81},$
$\text{P(X}=1)=\frac{\text{ }^4\text{C}_12^3}{3^4}=\frac{32}{81},$
$\text{P(X}=2)=\frac{\text{ }^4\text{C}_22^2}{3^4}=\frac{24}{81},$
$\text{P(X}=3)=\frac{\text{ }^4\text{C}_32^1}{3^4}=\frac{8}{81},$
$\text{P(X}=4)=\frac{\text{ }^4\text{C}_42^0}{3^4}=\frac{1}{81}$
So, the probability distribution of X is given as follows:
$\text{X}:$
 $0$
 $1$
 $2$
 $3$
 $4$
$\text{P(X):}$
 $\frac{16}{81}$
 $\frac{32}{81}$
 $\frac{24}{81}$
 $\frac{8}{81}$
 $\frac{1}{81}$
Now,
$\text{Mean, E(X)}=0\times\frac{16}{81}+1\times\frac{32}{81}+2\times\frac{24}{81}+3\times\frac{8}{81}+4\times\frac{1}{81}$
$=\frac{32+48+24+4}{81}$
$=\frac{108}{81}$
$=\frac{4}{3}$
Note: we can also calculate the mean of the binomial distribution by
$\text{Mean, E(X) = np}=4\times\frac{1}{3}=\frac{4}{3}$
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Question 274 Marks
The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?
Answer
Let hitting the target be a success in a shoot.
We have,
p = probability of hitting the target $=0.25=\frac{1}{4}$
Also, $\text{q}=1-\text{p}=1=\frac{1}{4}=\frac{3}{4}$
Let X denote the number of success in a sample of 7 trils. then,
X follows binomial distribution with parameters n = 7 and $\text{p}=\frac{1}{4}$
$\therefore\text{P(X = r})=\text{ }^7\text{C}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(7-\text{r})}=\text{ }^7\text{C}_{\text{r}}\big(\frac{1}{4}\big)^{\text{r}}\big(\frac{3}{4}\big)^{(7-\text{r})}=\frac{\text{ }^7\text{C}_{\text{r}}3^{(7-\text{r})}}{4^7},$ where r = 0, 1, 2, 3, 4, 5
Now,
Required probability $=\text{P(X}\geq2)$
$=1-\big[\text{P(X}=0)+\text{P(X}=1)\big]$
$=1-\Big[\frac{\text{ }^7\text{C}_03^7}{4^7}+\frac{\text{ }^7\text{C}_13^6}{4^7}\Big]$
$=1-\Big[\frac{2187}{16384}+\frac{5103}{16384}\Big]$
$=1-\frac{7290}{16384}$
$=\frac{9094}{16384}$
$=\frac{4547}{8192}$
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Question 284 Marks
From a lot of 30 bulbs that includes 6 defective bulbs, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Answer
Let X denote the number of defective bulls in a sample of 4 bulbs drawn successively with replacement.
Then, X follows a binomial distribution with the following parameters: n = 4,
$\text{p}=\frac{6}{30}=\frac{1}{5}$ and $\text{q}=\frac{4}{5}$
Then, the distribution is given by
$\text{P(X = r})=\text{ }^4\text{C}_{\text{r}}\big(\frac{1}{5}\big)^{\text{r}}\big(\frac{4}{5}\big)^{4-\text{r}}.\text{r}=0,1,2,3,4$
$\text{P(X = } 0)=\big(\frac{4}{5}\big)^4$
$=\frac{256}{625}$
$\text{P(X}=1)=4\big(\frac{1}{5}\big)^1\big(\frac{4}{5}\big)^3$
$=\frac{256}{625}$
$\text{P(X}=2)=6\big(\frac{1}{5}\big)^2\big(\frac{4}{5}\big)^2$
$=\frac{96}{625}$
$\text{P(X}=3)=4\big(\frac{1}{5}\big)^3\big(\frac{4}{5}\big)^1$
$=\frac{16}{625}$
$\text{P(X}=4)=\big(\frac{1}{5}\big)^4$
$=\frac{1}{625}$
$\text{X}$
 $1$
 $2$
 $3$
 $4$
 $5$
$1\text{P(X)}$
 $\frac{256}{625}$
 $\frac{256}{625}$
 $\frac{96}{625}$
 $\frac{16}{625}$
 $\frac{1}{625}$
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Question 294 Marks
Assume that the probability that a bomb dropped from an aeroplane will strike a certain target is 0.2. If 6 bombs are dropped, find the probability that.
  1. exactly 2 will strike the target.
  2. at least 2 will strike the target.
Answer
probabilty that bomb strikes a target p = 0.2
Probability that a bomb misses the target = 0.8
n = 6
let X = number of bombs that strike the target
P(X = 2) = exactly 2 bombs strike the target
$=\text{ }^6\text{C}_2\big(\frac{2}{10}\big)^2\times\big(\frac{8}{10}\big)^4=15\times\frac{16384}{10^6}=0.24576$
$\text{P(X}\geq2)=$ at least 2 bombs strike the target
$=1-\text{P(X}<2)$
$=1-\big[\text{P(X}=0)+\text{P(X=1})\big]$
$=1-\big[\text{ }^6\text{C}_0\big(\frac{2}{10}\big)^0\times\big(\frac{8}{10}\big)^6+\text{ }^6\text{C}_1\big(\frac{2}{10}\big)^1\times\big(\frac{8}{10}\big)^6\big]$
$=1-\big[0.0.262144+0.393216\big]=1-0.65536$
$=0.34464$
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Question 304 Marks
How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?
Answer
Suppose the man tosses a fair coin n times and X denote the number of heads in n tosses.
As $\text{p}=\frac{1}{2}$ and $\text{q}=\frac{1}{2},$
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{\text{n}-\text{r}},\text{r}=0,1,2,3\dots\text{n}$
It is given that $\text{P(X}\geq1)>0.9$
$\Rightarrow1-\text{P(X}=0)>0.9$
$\Rightarrow1-\text{ }^{\text{n}}\text{C}_0\big(\frac{1}{2}\big)^{\text{n}}>0.9$
$\Rightarrow\big(\frac{1}{2}\big)^{\text{n}}<\frac{1}{10}$
$\Rightarrow2^{\text{n}}>10$
$\Rightarrow\text{n}=4,5,6\dots$
Hence, the man must tosss the coin at least 4 times.
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Question 314 Marks
In a multiple-choice examination with three possible answers for each of the five questions out of which only one is correct, what is the probability that a candidate would get four or more correct answers just by guessing?
Answer
The repeated guessing of correct answers form 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 $\text{n = 5}$ and $\text{p}=\frac{1}{3}$

$\therefore\text{P(X = x)}=\text{ }^{\text{n}}\text{C}_{\text{x}}\text{q}^{\text{n}-\text{x}}\text{p}^{\text{x}}$

$=\text{ }^5\text{C}_{\text{x}}\big(\frac{2}{3}\big)^{5-\text{x}}.\big(\frac{1}{3}\big)^{\text{x}}$

P(guessing more than 4 correct answer) $=\text{P(X}\geq4)$

$=\text{P(X}=4)+\text{P(X}=5)$

$=\text{ }^5\text{C}_4\big(\frac{2}{3}\big).\big(\frac{1}{3}\big)^4+\text{ }^5\text{C}_5\big(\frac{1}{3}\big)^5$

$=5.\frac{2}{3}.\frac{1}{81}+1.\frac{1}{243}$

$=\frac{10}{243}+\frac{1}{243}$

$=\frac{11}{243}$

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Question 324 Marks
A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.
Answer
Let geting an odd number be a success in trial.
We have,
p = probability of getting an odd number in a trial $=\frac{3}{6}=\frac{1}{2}$
Also, $\text{q}=1-\text{p}=1-\frac{1}{2}=\frac{1}{2}$
Let X denote the number of success in a sample of 5 trials. Then,
X follows binomial distribution with parameters n = 5 and $\text{p = q}=\frac{1}{2}$
$\therefore\text{P(X = r})=\text{ }^{5}\text{C}_{\text{r}}\text{p}^{\text{r}}\text{q}^{(5-\text{r})}=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{(5-\text{r})}=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^5,$ wher r = 0, 1, 2, 3, 4, 5
Now,
Required probability = P(X = 3)
$=\text{ }^5\text{C}_3\big(\frac{1}{2}\big)^5$
$=\frac{10}{32}$
$=\frac{5}{16}$
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