Solve the Following Question.(5 Marks)

Question 15 Marks

In a large school, $80\%$ of the pupil like Mathematics. A visitor to the school asks each of $4$ pupils, chosen at random, whether they like Mathematics.
(i) Calculate the probabilities of obtaining an answer yes from$ 0, 1, 2, 3, 4$ of the pupils.
(ii) Find the probability that the visitor obtains answer yes from at least$ 2$ pupils:
(a) when the number of pupils questioned remains at $4.$
(b) when the number of pupils questioned is increased to $8.$

Answer

Let $X=$ number of pupils like Mathematics.
$p=$ probability that pupils like Mathematics
$ \therefore p=80 \%=\frac{80}{100}=\frac{4}{5}$
$\text { and } q=1-p=1-\frac{4}{5}=\frac{1}{5} $
Given : $n=4$
$\therefore x \sim B\left(4, \frac{4}{5}\right)$
The p.m.f. of $X$ is given by
$P(X=x)={ }^n \mathrm{C}_x P^x q^{n-x}$
i.e. $p(x)={ }^4 \mathrm{C}_x\left(\frac{4}{5}\right)^x\left(\frac{1}{5}\right)^{4-x}, x=0,1,2,3,4$
(i) The probabilities of obtaining an answer yes from $0,1,2,3,4$ of pupils are $P(X=$ $P(X=1), P(X=2), P(X=3)$ and $P(X=4)$ respectively
i.e. ${ }^4 C_0\left(\frac{4}{5}\right)^0\left(\frac{1}{5}\right)^{4-0},{ }^4 C_1\left(\frac{4}{5}\right)^1\left(\frac{1}{5}\right)^{4-1}$,
${ }^4 \mathrm{C}_2\left(\frac{4}{5}\right)^2\left(\frac{1}{5}\right)^{4-2},{ }^4 \mathrm{C}_3\left(\frac{4}{5}\right)^3\left(\frac{1}{5}\right)^{4-3} \text { and }$
$^4 \mathrm{C}_4\left(\frac{4}{5}\right)^4\left(\frac{1}{5}\right)^{4-4}$
i.e. $1(1)\left(\frac{1}{5}\right)^4, 4\left(\frac{4}{5}\right) \cdot\left(\frac{1}{5}\right)^3, \frac{4 \times 3}{1 \times 2}\left(\frac{16}{25}\right)\left(\frac{1}{25}\right)$,
$4\left(\frac{64}{125}\right)\left(\frac{1}{5}\right) \text { and } 1 \times\left(\frac{4}{5}\right)^4\left(\frac{1}{5}\right)^0$
i.e. $\left(\frac{1}{5}\right)^4, \frac{16}{5}\left(\frac{1}{5}\right)^3, \frac{96}{5^2}\left(\frac{1}{5^2}\right), \frac{256}{5^3}\left(\frac{1}{5}\right)$ and $\frac{256}{5^4}$
i.e. $\frac{1}{5^4}, \frac{16}{5^4}, \frac{96}{5^4}, \frac{256}{5^4}, \frac{256}{5^4}$.
$O R \frac{1}{625}, \frac{16}{625}, \frac{96}{625}, \frac{256}{625}$ and $\frac{256}{625}$.
(ii) (a) P(visitor obtains the answer yes from at least 2 pupils when the number of pupils questioned remains at 4$)=P(X \geq 2)$
$ =P(X=2)+P(X=3)+P(X=4)$
$={ }^4 C_2\left(\frac{4}{5}\right)^2\left(\frac{1}{5}\right)^2+{ }^4 C_3\left(\frac{4}{5}\right)^3\left(\frac{1}{5}\right)^1+{ }^4 C_4\left(\frac{4}{5}\right)^4\left(\frac{1}{5}\right)^0$
$=\frac{4 \times 3}{1 \times 2} \times \frac{16}{5^2} \times \frac{1}{5^2}+4 \times \frac{64}{5^3} \times \frac{1}{5}+1 \times \frac{256}{5^4}$
$=\frac{96}{5^4}+\frac{256}{5^4}+\frac{256}{5^4}$
$=(96+256+256) \frac{1}{5^4}$
$=\frac{608}{5^4}=\frac{608}{625} . $
(b) $P$ (the visitor obtains the answer yes from at least 2 pupils when number of pupils questioned is increased to 8 )
$ =P(X \geqslant 2)$
$=1-P(X<2)$
$=1-[P(X=0)+P(X=1)]$
$=1-\left[{ }^8 C_0\left(\frac{4}{5}\right)^0\left(\frac{1}{5}\right)^{8-0}+{ }^8 C_1\left(\frac{4}{5}\right)^1\left(\frac{1}{5}\right)^{8-1}\right]$
$=1-\left[1(1)\left(\frac{1}{5}\right)^8+(8)\left(\frac{4}{5}\right)\left(\frac{1}{5}\right)^7\right]$
$=1-\left[\frac{1}{5^8}+\frac{32}{5^8}\right]$
$=1-\frac{33}{5^8} . $

View full question & answer→

Question 25 Marks

The probability that a mountain bike travelling along a certain track will have a tire burst is $0.05.$ Find the probability that among $17$ riders:
(i) exactly one has a burst tyre
(ii) at most three have a burst tyre
(iii) two or more have burst tyres.

Answer

Let $X=$ number of burst tyres.
$p=$ probability that a mountain bike travelling along a certain track will have a tyre burst.
$ \therefore p=0.05$
$\therefore q=1-p=1-0.05=0.95 $
Given: $n=17$
$\therefore X \sim B(17,0.05)$
The p.m.f. of $X$ is given by
$ \mathrm{P}(\mathrm{X}=\mathrm{x})={ }^n \mathrm{C}_x P^x q^{n-x}$
$\text { i.e. }(\mathrm{x})={ }^{17} \mathrm{C}_x(0.05)^x(0.95)^{17-x}, \mathrm{x}=0,1,2, \ldots \ldots, 17 $
(i) P(exactly one has a burst tyre)
$ P(X=1)=p(1)={ }^{17} C_1(0.05)^1(0.95)^{17-1}$
$=17(0.05)(0.95)^{16}$
$=0.85(0.95)^{16} $
Hence, the probability that riders has exactly one burst tyre $=(0.85)(0.95)^{16}$
$ \text { (ii) } P \text { (at most three have a burst tyre) }=P(X \leq 3)$
$=P(X=0)+P(X=1)+P(X=2)+P(X=3)$
$=p(0)+p(1)+p(2)+p(3)$
$={ }^{17} \mathrm{C}_0(0.05)^0(0.95)^{17-0}+{ }^{17} \mathrm{C}_1(0.05)^1(0.17)^{17-1}+$
${ }^{17} C_2(0.05)^2(0.17)^{17-2}+{ }^{17} C_3(0.05)^3(0.17)^{17-3}$
$=1(1)(0.95)^{17}+17(0.05)(0.95)^{16}+$
$\frac{17 \times 16}{2 \times 1} \times(0.05)^2(0.95)^{15}+$
$\frac{17 \times 16 \times 15}{3 \times 2 \times 1} \times(0.05)^3 \times(0.95)^{14}$
$=(0.95)^{17}+17(0.05) \times(0.95)^{16}+$
$17(8) \times(0.05)^2 \times(0.95)^{15}+17(8)(5) \times(0.05)^3 \times(0.95)^{14}$
$=(0.95)^{14}\left[(0.95)^3+(17)(0.05)(0.95)^2+\right.$
$\left.17(8) \times(0.05)^2 \times(0.95)^1+17(8)(5)(0.05)^3\right]$
$=(0.95)^{14}[0.8574+0.7671+0.323+0.085]$
$=(2.0325)(0.95)^{14}$
Hence, the probability that at most three riders have burst tyre $=(2.0325)(0.95)^{14}$.
$ \text { (iii) } P \text { (two or more have tyre burst) }=P(X \geq 2)$
$=1-P(X<2)$
$=1-[P(X=0)+P(X=1)]$
$=1-[P(0)+p(1)]$
$=1-\left[{ }^{17} C_0(0.05)^0(0.95)^{17}+{ }^{17} C_1(0.05)(0.95)^{16}\right]$
$=1-\left[1(1)(0.95)^{17}+17(0.05)(0.95)^{16}\right]$
$=1-(0.95)^{16}[0.95+0.85]$
$=1-(1.80)(0.95)^{16}$
$=1-(1.8)(0.95)^{16} $
Hence, the probability that two or more riders have tyre burst $=1-(1.8)(0.95)^{16}$.

View full question & answer→

Question 35 Marks

In a box of floppy discs, it is known that 95% will work. A sample of three of the discs is selected at random. Find the probability that (i) none (ii) 1 (iii) 2 (iv) all 3 of the sample will work.

Answer

Let $X=$ number of working discs.
$\mathrm{p}=$ probability that a floppy disc works
$\therefore p=95 \%=\frac{95}{100}=\frac{19}{20}$
and $q=1-p=1-\frac{19}{20}=\frac{1}{20}$
Given: $\mathrm{n}=3$
$\therefore X \sim \mathrm{B}\left(3, \frac{19}{20}\right)$
The p.m.f. of $X$ is given by
$P(X=x)={ }^n C_x p^x q^{n-x}$
i.e. $p(x)={ }^3 C_x\left(\frac{19}{20}\right)^x\left(\frac{1}{20}\right)^{3-x}, x=0,1,2,3$
(i) $P$ (none of the floppy discs work) $=P(X=0)$
$=p(0)={ }^3 \mathrm{C}_0\left(\frac{19}{20}\right)^0\left(\frac{1}{20}\right)^{3-0}$
$=1 \times 1 \times \frac{1}{20^3}=\frac{1}{20^3}$
Hence, the probability that none of the floppy disc will work $=\frac{1}{20^3}$.
(ii) $P$ (exactly one floppy disc works) $=P(X=1)$
$ =p(1)={ }^3 \mathrm{C}_1\left(\frac{19}{20}\right)^1\left(\frac{1}{20}\right)^{3-1}$
$=3 \times \frac{19}{20} \times\left(\frac{1}{20}\right)^2$
$=3\left(\frac{19}{20^3}\right) $
$=3\left(\frac{19}{20^3}\right)$
Hence, the probability that exactly one floppy disc works $=3\left(\frac{19}{20^3}\right)$
(iii) $\mathrm{P}$ (exactly two floppy discs work $)=P(X=2)$
$=p(2) ={ }^3 C_2\left(\frac{19}{20}\right)^2\left(\frac{1}{20}\right)^{3-2}$
$ =\frac{3 \cdot 2 !}{2 ! \cdot 1 !} \times \frac{(19)^2}{(20)^2} \times \frac{1}{20}=3\left(\frac{19^2}{20^3}\right) $
Hence, the probability that exactly 2 floppy discs work $=3\left(\frac{19^2}{20^3}\right)$
(iv) $P$ (all 3 floppy discs work $)=P(X=3)$
$=p(3) ={ }^3 C_3\left(\frac{19}{20}\right)^3\left(\frac{1}{20}\right)^{3-3}$
$ =1 \times\left(\frac{19}{20}\right)^3 \times\left(\frac{1}{20}\right)^0$
$ =\left(\frac{19}{20}\right)^3 $
Hence, the probability that all 3 floppy discs work $=\left(\frac{19}{20}\right)^3$.

View full question & answer→

Question 45 Marks

The probability of a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one, will fuse after 150 days of use.

Answer

Let $X=$ number of fuse bulbs.
$p=$ probability of a bulb produced by a factory will fuse after 150 days of use.
$\therefore p =0.05$
and $q=1-p=1-0.05=0.95$
Given: $n =5$
$\therefore X \sim B (5,0.05)$
The p.m.f. of $X$ is given by
$ P ( X = x )={ }^n C _x p^x q^{n-x}$
$\text { i.e. } p ( x )={ }^5 C_x(0.05)^x(0.95)^{5-x}, x =0,1,2,3,4,5 $
(i) $P$ (none of a bulb produced by a factory will fuse after 150 days of use) $=P[X=0]$
$ = p (0)$
$={ }^5 C _0(0.05)^0(0.95)^{5-0}$
$=1 \times 1 \times(0.95)^5$
$=(0.95)^5 $
Hence, the probability that none of the bulbs will fuse after 150 days $=(0.95)^5$.
(ii) $P$ (not more than one bulb will fuse after 150 days of $j$ use) $=P[X \leq 1]$
$ =p(0)+p(1)$
$={ }^5 C _0 \cdot(0.05)^0(0.95)^{5-0}+{ }^5 C _1(0.05)^1(0.95)^4$
$=1 \times 1 \times(0.95)^5+5 \times(0.05) \times(0.95)^4$
$=(0.95)^4[0.95+5(0.05)]$
$=(0.95)^4(0.95+0.25)$
$=(0.95)^4(1.20)$
$=(1.2)(0.95)^4 $
Hence, the probability that not more than one bulb will fuse after 150 days $=(1.2)(0.95)^4$.
(iii) $P$ (more than one bulb fuse after 150 days)
$ =P[X>1]$
$=1-P[X \leq 1]$
$=1-(1.2)(0.95)^4 $
Hence, the probability that more than one bulb fuse after 150 days $=1-(1.2)(0.95)^4$.
(iv) $P$ (at least one bulb fuse after 150 days)
$ =P[X \geq 1]$
$=1-P[X=0]$
$=1-p(0)$
$=1-{ }^5 C_0(0.05)^0(0.95)^{5-0}$
$=1-1 \times 1 \times(0.95)^5$
$=1-(0.95)^5 $
Hence, the probability that at least one bulb fuses after 150 days $=1-(0.95)^5$.

View full question & answer→

Question 55 Marks

Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards, find the probability that
(i) all the five cards are spades
(ii) only 3 cards are spades
(iii) none is a spade.

View full question & answer→

Question 65 Marks

There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?

View full question & answer→

Question 75 Marks

A die is thrown 6 times. If ‘getting an odd number’ is a success, find the probability of
(i) 5 successes
(ii) at least 5 successes
(iii) at most 5 successes.

View full question & answer→

Maths Solve the Following Question.(5 Marks)

Take a timed test