Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

A.  $$\frac{9}{20}$$
B.  $$\frac{7}{3}$$
C.  $$\frac{3}{7}$$
D.  $$\frac{8}{7}$$

Solution:

$\left. \begin{array} { l } { \text { Here, } S = \{ 1,2,3,4 , \ldots , 19,20 \} \text { . } } \\ { \text { Let } E = \text { event of getting a multiple of } 3 \text { or } 5 }\\ {= \{ 3,6,9,12,15,18,5,10,20 \} } \\ { \therefore P ( E ) = \frac { n ( E ) } { n ( S ) } = \frac { 9 } { 20 } } \end{array} \right.$

In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

A.  $$\frac{4}{3}$$
B.  $$\frac{1}{3}$$
C.  $$\frac{2}{5}$$
D.  $$\frac{5}{3}$$

Solution:

$\text { Total number of balls } = ( 8 + 7 + 6 ) = 21$ Let E = event that the ball drawn is neither red nor green
$\left. \begin{array} { l } { \quad = \text { event that the ball drawn is blue. } } \\ { n ( E ) = 7 } \\ { P ( E ) = \frac { n ( E ) } { n ( S ) } = \frac { 7 } { 21 } = \frac { 1 } { 3 } } \end{array} \right.$

What is the probability of getting a sum 9 from two throws of a dice?

A.  $$\frac{5}{7}$$
B.  $$\frac{3}{2}$$
C.  $$\frac{2}{3}$$
D.  $$\frac{1}{9}$$

Solution:

$\left. \begin{array} { l } { \text { In two throws of a dice, } n ( S ): }\\ {= ( 6 \times 6 ) = 36 \text { . } } \\ { \text { Let } E = \text { event of getting a sum } = \{ ( 3,6 ) ,}\\ { ( 4,5 ) , ( 5,4 ) , ( 6,3 ) \} \text { . } } \\ { \therefore P ( E ) = \frac { n ( E ) } { n ( S ) } = \frac { 4 } { 36 } = \frac { 1 } { 9 } } \end{array} \right.$

Three unbiased coins are tossed. What is the probability of getting at most two heads?

A.  $$\frac{7}{8}$$
B.  $$\frac{2}{7}$$
C.  $$\frac{3}{5}$$
D.  $$\frac{1}{3}$$

Solution:

$\left. \begin{array} { l } { \text { Here } S = \{ TTT , TTH , THT , HTT ,}\\ { THH , HTH , HHT , HHH \} } \\ { \text { Let } E = \text { event of getting at most two heads. } } \\ { \text { Then } E = \{ TTT , TTH , THT , HTT , THH ,}\\ { HTH , HHT \} } \\ { \therefore P ( E ) = \frac { n ( E ) } { n ( S ) } = \frac { 7 } { 8 } } \end{array} \right.$

Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

A.  $$\frac{3}{52}$$
B.  $$\frac{3}{7}$$
C.  $$\frac{3}{12}$$
D.  $$\frac{3}{4}$$

Solution:

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
\begin{array} { l } { \text { Then, } E = \{ ( 1,2 ) , ( 1,4 ) , ( 1,6 ) , ( 2,1 ) , ( 2,2 )}\\
{ , ( 2,3 ) , ( 2,4 ) , ( 2,5 ) , ( 2,6 ) , ( 3,2 ) , ( 3,4 ) , } \\ { ( 3,6 ) , ( 4,1 ) , ( 4,2 ) , ( 4,3 ) , ( 4,4 ) , ( 4,5 ) , ( 4,6 )}\\
{ , ( 5,2 ) , ( 5,4 ) , ( 5,6 ) , ( 6,1 ) , } \\ { \quad ( 6,2 ) , ( 6,3 ) , ( 6,4 ) , ( 6,5 ) , ( 6,6 ) \} } \\ { \therefore n ( E ) = 27 } \\ { \therefore P ( E ) = \frac { n ( E ) } { n ( S ) } = \frac { 27 } { 36 } = \frac { 3 } { 4 } } \end{array}

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

A.  $$\frac{21}{46}$$
B.  $$\frac{12}{43}$$
C.  $$\frac{2}{13}$$
D.  $$\frac{53}{34}$$

Solution:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
$\left. \begin{array} { l } { \left. \begin{array}{l}{ \text { Then, } n ( S ) = \text { Number ways of selecting }}\\ { 3 \text { students out of } 25 }\\{ = 25 C _ { 3 } }\\{ = \frac { ( 25 \times 24 \times 23 ) } { ( 3 \times 2 \times 1 ) } }\\{ = 2300 }\end{array} \right. } \\ { \left. \begin{array}{l}{ = \frac { 125 \times 10 } { ( 2 \times 1 ) } }\\{ \therefore 100 C _ { 1 } \times ^ { 15 } C _ { 2 } ) }\\{ = [ 10 \times \frac { ( 15 \times 14 ) } { ( 2 \times 1 ) } ] }\end{array} \right. } \\ { \left. \begin{array}{l}{ n ( E ) = 1050 }\\{ \therefore P ( E ) = \frac { n ( E ) } { n ( S ) } = \frac { 1050 } { 2300 } = \frac { 21 } { 46 } }\end{array} \right. } \end{array} \right.$

In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

A.  $$\frac{4}{3}$$
B.  $$\frac{7}{6}$$
C.  $$\frac{3}{4}$$
D.  $$\frac{2}{7}$$

Solution:

$P ( \text { getting a prize } ) = \frac { 10 } { ( 10 + 25 ) } = \frac { 10 } { 35 } = \frac { 2 } { 7 }$

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