MATHS 5 Marks Questions

7-White balls

5-Black balls

3-Red balls

Since two balls are drawn at random

$\therefore\text{n}(\text{S})=\frac{15}{2}$

1. E be the event that both the balls are red

$\therefore\text{n}(\text{S})=​​^3\text{C}_2$

$\therefore\text{p}(\text{E})=\frac{^3\text{C}_2}{^{15}\text{C}_2}=\frac{3\times2}{15\times14}=\frac{1}{35}$

2. E be the event that one ball is red and other is black

$\therefore\text{n}(\text{E})=^3\text{C}_1\times^5\text{C}_1$​​​​​​​

$\therefore\text{p}(\text{E})=\frac{^3\text{C}_1\times^5\text{C}_1}{^{15}\text{C}_2}$​​​​​​​

$=\frac{3\times5\times2}{15\times16}=\frac{1}{7}$

3. E be the event that one ball is white

$\therefore\text{n}(\text{E})=^7\text{C}_1\times^8\text{C}_1$

$\therefore\text{p}(\text{E})=\frac{^7\text{C}_1\times^8\text{C}_1}{^{15}\text{C}_2}$

$=\frac{7\times6\times2}{14\times15}=\frac{8}{15}$

1. All 5 are blue

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

2. 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

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

$=\frac{4367}{4484}$

1. $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$

$=0.35+0.45-0$

$=0.80$

2. A and B are mutually exclusive events.

​​​​​​​$\therefore\text{P}({\text{A}}\cap{\text{B}})=0$

3. $\text{P}({\text{A}}\cap\overline{\text{B}})=\text{p}(\text{A})-\text{P}({\text{A}}\cap{\text{B}})$

$=0.35-0$

$=0.35$

4. $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup{\text{B}}})$

$=1-\text{P}({\text{A}\cup{\text{B}}})$

$=1-0.80$

$=0.20$

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

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

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

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

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

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

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

3. 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

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

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

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

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

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

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

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

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

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

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

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

2. 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

6. 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

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

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

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

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

1. E be the event that the group has all boys

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

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

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

$=\frac{5}{34}$

2. E be the event that the group has all girls

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

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

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

$=\frac{7}{102}$

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

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

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

$=\frac{35}{102}$

4. E be the event that atleast one girls in the group

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

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

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

$=\frac{29}{34}$

5. E be the event that almost one girls in the group

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

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

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

$=\frac{10}{17}$

1. Let E be the event that those five card contain exactly one ace.

$\therefore\text{n}(\text{E})=\ ^{4}\text{C}_1\times\ ^{48}\text{C}_4$

$\therefore\text{n}(\text{E})=\frac{^4\text{C}_1\times^{48}\text{C}_4}{^{52}\text{C}_5}$

$=\frac{4\times48\times47\times46\times45}{\frac{52\times51\times50\times49\times48}{5}}$

$=\frac{3243}{10829}$

2. Let E be the event that five cards contain atleast one  ace.

$\therefore\text{E}= \big\{{1\ \text{or}\ 2\text{ or } 3 \text{ or }4}\big\}$

$\text{n}(\text{E})=\frac{^4\text{C}_1\times^{48}\text{C}_4\ +\ ^4\text{C}_2\times^{48}\text{C}_3\ +\ ^4\text{C}_3\times^{48}\text{C}_2\ +\ ^4\text{C}_4\times^{48}\text{C}_1}{^{52}\text{C}_5}$

$=\frac{4\times\frac{48\times47\times46\times45}{4\times3\times2\times1}+\frac{4\times3}{2}\times\frac{48\times47\times46}{3\times2\times1}+4\times\frac{48\times47}{2}+48}{\frac{52\times51\times50\times49\times48}{5\times4\times3\times2\times1}}$

$=\frac{18472}{54145}$