Solve the Following Question.(5 Marks)

Question 15 Marks

A bag contains 6 red, 4 white and 8 blue balls. if three balls are drewn at random, find the probability that:

(i) One is red and two are white

(ii) Two are blue and one is red

(iii) One is red.

Answer

BAG: 6 - Red ball, 4 - White ball, 8 - blue ball
Since three ball are drawn,
$\therefore\text{n}\text{(S)}=\ ^{18}\text{C}_3$
(i) Let E be the events that one red and two white ball are drawn.
$\therefore\text{n}\text{(S)}=\ ^{6}\text{C}_1\times\ ^{4}\text{C}_2$
$\therefore\text{P}\text{(E)}=\frac{\ ^{6}\text{C}_1\times\ ^{4}\text{C}_2}{\ ^{18}\text{C}_3}=\frac{6\times4\times3}2\times{\frac{3\times2}{18\times17\times16}}$
$\text{P}\text{(E)}=\frac{3}{68}$
(ii) Let E be the event that two blue and one red ball are drawn.
$\therefore\text{n}\text{(E)}=\ ^{8}\text{C}_2\times\ ^{6}\text{C}_1$
$\therefore\text{P}\text{(E)}=\frac{\ ^{8}\text{C}_2\times\ ^{4}\text{C}_2}{\ ^{18}\text{C}_3}=\frac{8\times7}2\times6\times{\frac{3\times2\times1}{18\times17\times16}}=\frac{7}{34}$
$\text{P}\text{(E)}=\frac{7}{34}$
(iii) Let E be the event thatone of the ball must be red.
$\therefore\text{E}=\big\{(\text{R, W, B)}\ \text{or}\ \big\{(\text{R, W, W)}\ \text{or}\ \big\{(\text{R, B, B)}$
$\therefore\text{n}\text{(E)}=\ ^6\text{C}_1\times\ ^4\text{C}_1\times \ ^8\text{C}_1+\ ^6\text{C}_1\times \ ^4\text{C}_2+\ ^6\text{C}_1\ \times\ ^8\text{C}_2$
$\therefore\text{P}\text{(E)}=\frac{\ ^6\text{C}_1\times\ ^4\text{C}_1\times \ ^8\text{C}_1+\ ^6\text{C}_1\times \ ^4\text{C}_2+\ ^6\text{C}_1\ \times\ ^8\text{C}_2}{\ ^{18}\text{C}_3}$
$=\frac{396}{816}=\frac{33}{68}$

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Question 25 Marks

A box contains 100 bulbs, 20 of which are defective. 10 bulbs are selected for inspection. find the probability that:

All 10 are defective
All 10 are good
At least one is defective
None is defective

Answer

A box contains 100 bulbs, out of which 20 are defective,
$\therefore$ Numbers of good bulbs 100 - 20 = 80
Now,
10 balls are selected for inspection
$\therefore$ Numbers of elementary events in sample space
$\text{n}\text{(S)}=\ ^{100}\text{C}_{10}$

Let E be the events that all 10 bulbs selected are defective

$\text{n}\text{(E)}=\ ^{20}\text{C}_{10}$

$\therefore\text{P}\text{(E)}=\frac{\ ^{20}\text{C}_{10}}{\ ^{100}\text{C}_{10}}$

$=\frac{\ ^{20}\text{C}_{10}}{\ ^{100}\text{C}_{10}}$

Let E be the events that all 10 bulbs good bulbs are selected

$\therefore\text{n}\text{(E)}=\ ^{80}\text{C}_{10}$

$\therefore\text{P}\text{(E)}=\frac{\ ^{80}\text{C}_{10}}{\ ^{100}\text{C}_{10}}$

Let E be the events thatatleast one bulbs is defective

$\text{E}=\big\{1,\ 2,\ 3,\ 4,\ 5, \ 6,\ 7,\ 8,\ 9,\ 10\big\}$

Where,

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, are numbers of defective bulbs

$\because\ \stackrel{{\sim}}{\hbox{E}}$ be the event that none of the bulbs are defective

$\therefore \text{n}\stackrel{{\sim}}{\hbox{(E)}}=\ ^{80}\text{C}_{10}$

$\therefore \text{P}\stackrel{{\sim}}{\hbox{(E)}}=\frac{\ ^{80}\text{C}_{10}}{\ ^{100}\text{C}_{10}}$

$\therefore\text{P}\text{(E)}=1-\text{P}\stackrel{{\sim}}{\hbox{(E)}}$

$=1-\frac{\ ^{80}\text{C}_{10}}{\ ^{100}\text{C}_{10}}$

Let E be the events thatatleast one bulbs is defective, that is all bulbs are good so,

$\text{n}\text{(E)}=\ ^{80}\text{C}_{10}$

$\text{P}\text{(E)}=\frac{\ ^{80}\text{C}_{10}}{\ ^{100}\text{C}_{10}}$

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Question 35 Marks

20 cards are numbered form 1 to 20. card is drawn at random. what is the probability that trhe number on the card is:

A multiple of 4?
Not a multiple of 4?
odd?
Greather than 12?
Divisible by 5?
Not a multiple of 6?

Answer

We have 20 cards numbered from 1 to 20, one card is drawn at random
$\therefore\text{n(S)}=\ ^{20}\text{C}_1=20$

Let E be the event that the number on the drawn cards is multiple of 4

$\therefore\text{E}=\big\{4, \ 8, \ 12,\ 16,\ 20\big\}$

$\therefore\text{n(E)}=5$

$\therefore\text{p(E)}=\frac{5}{20}=\frac{1}{4}$

Let E be the event that the number on the drawn cards is not multiple of 4

$\therefore\ \stackrel{{\sim}}{\hbox{E}}$ be the event that the number on the drawn cards is not multiple of 4

$\therefore\ \stackrel{{\sim}}{\hbox{E}}\ =\big\{4,\ 8, \ 12, \ 16,\ 20\big\}$

$\Rightarrow\text{n}\stackrel{{\sim}}{\hbox{(E)}}=5$

$\therefore\text{P}\stackrel{{\sim}}{\hbox{(E)}}=\frac{5}{20}=\frac{1}{4}$

$\therefore\text{P(E)}=1-\text{P}(\stackrel{{\sim}}{\hbox{E)}}$

$=1-\frac{1}{4}=\frac{3}{4}$

Let E be the event that the number on the drawn cards is odd.

$\therefore\text{E}=\big\{1, \ 3,\ 5,\ 7,\ 13,\ 15,\ 17,\ 19\big\}$

$\therefore\text{n(E)}=10$

$\Rightarrow\text{P(E)}=\frac{10}{20}=\frac{1}{2}$

Let E be the event that the number on the drawn cards is greater than 12.

$\therefore\text{E}=\big\{13, \ 14,\ 15,\ 16,\ 17,\ 18,\ 19,\ 20\big\}$

$\therefore\text{n(E)}=8$

$\Rightarrow\text{P(E)}=\frac{8}{20}=\frac{2}{5}$

Let E be the event that the number on the drawn cards is divisible by 5.

$\therefore\text{E}=\big\{5, \ 10,\ 15,\ 20\big\}$

$\text{n(E)}=4$

$\therefore\text{P(E)}=\frac{4}{20}=\frac{1}{5}$

Let E be the event that the number on the drawn cards is divisible by 6.

$\therefore\ \stackrel{{\sim}}{\hbox{E}}$ be the event that the number on the drawn cards is not divisible of 6

$\Rightarrow\text{n}\stackrel{{\sim}}{\hbox{(E)}}=3$

$\therefore\text{P}\stackrel{{\sim}}{\hbox{(E)}}=\frac{3}{20}$

$\text{P(E)}=1-\text{P}(\stackrel{{\sim}}{\hbox{E)}}$

$=1-\frac{3}{20}=\frac{17}{20}$

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Question 45 Marks

A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn at random. from the box, what is the probability that:

All are blue?
At least one is green?

Answer

10 Red, 20 Blue, 30 Green

All 5 are blue

$=\frac{\ ^{20}\text{C}_5\times\ ^{40}\text{C}_0}{\ ^{60}\text{C}_5}$ $=\frac{34}{11977}$

Atleast one green = 1 - no green

Different combination one possible for no green case are 5B, 1R, 2R, 3B, 3R, 2B, 4R, 2B, 5R,

$5\text{B}=\ ^{20}\text{C}_5$

$1\text{R}\ 4\text{B}=\ ^{10}\text{C}_1\times\ ^{20}\text{C}_4$

$2\text{R}\ 3\text{B}=\ ^{10}\text{C}_2\times\ ^{20}\text{C}_3$

$3\text{R}\ 2\text{B}=\ ^{10}\text{C}_3\times\ ^{20}\text{C}_2$

$4\text{R}\ 1\text{B}=\ ^{10}\text{C}_1\times\ ^{20}\text{C}_1$

$5\text{R}=\ ^{10}\text{C}_5$

Atleast one green = 1 - no green

$=1-\frac{\ ^{20}\text{C}_5+\ ^{10}\text{C}_1\times\ ^{20}\text{C}_4+\ ^{10}\text{C}_2\times\ ^{20}\text{C}_3+\ ^{10}\text{C}_3\times\ ^{20}\text{C}_2+\ ^{10}\text{C}_4\times\ ^{20}\text{C}_1+\ ^{10}\text{C}_5}{\ ^{60}\text{C}_5}$

$=\frac{4367}{4484}$

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Question 55 Marks

A class consists of 10 boys and 8 girls. Three students are selected at random. What is the probability that the selected group has

All boys?
All girls?
1 boys and 2 girls?
At least one girl?
At most one girl?

Answer

10 boys
8 girls
Three student are selected at random
$\text{n}(\text{S})=^{18}\text{C}_3$

E be the event that the group has all boys

$\therefore\text{n}(\text{E})=^{18}\text{C}_3$

$\therefore\text{p}(\text{E})=\frac{^{10}\text{C}_3}{^{18}\text{C}_3}$

$=\frac{10\times9\times8}{18\times17\times16}$

$=\frac{5}{34}$

E be the event that the group has all girls

$\therefore\text{n}(\text{E})=^8\text{C}_3$

$\therefore\text{p}(\text{E})=\frac{^{8}\text{C}_3}{^{18}\text{C}_3}$

$=\frac{8\times7\times6}{18\times17\times16}$

$=\frac{7}{102}$

E be the event that the group has one boy and two girls

$\therefore\text{n}(\text{E})=^8\text{C}_3\times^{10}\text{C}_2$

$\therefore\text{p}(\text{E})=\frac{^{8}\text{C}_1\times^{10}\text{C}_2}{^{18}\text{C}_3}$

$=\frac{35}{102}$

E be the event that atleast one girls in the group

$\text{E}=(1,\ 2,\ 3)\text{girls}$

$\therefore\text{n}(\text{E})=^8\text{C}_1\times^{10}\text{C}_2+^8\text{C}_2\times^{10}\text{C}_1+^8\text{C}_3\times^{10}\text{C}_0$

$\text{p}(\text{E})=\frac{^{8}\text{C}_1\times^{10}\text{C}_2+^{8}\text{C}_2\times^{10}\text{C}_1+^{8}\text{C}_3}{^{18}\text{C}_3}$

$=\frac{29}{34}$

E be the event that almost one girls in the group

$\text{E}=(0,\ 1,)\text{girls}$

$\therefore\text{n}(\text{E})=^8\text{C}_0\times^{10}\text{C}_3+^8\text{C}_1\times^{10}\text{C}_2$

$\text{p}(\text{E})=\frac{^{10}\text{C}_3+8\times^{10}\text{C}_2}{^{18}\text{C}_3}$

$=\frac{10}{17}$

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Maths Solve the Following Question.(5 Marks)

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