Blog

Parin Sir > Quantitative Aptitude (Maths) > Probability > Examples 5 to 8 > Examples 5 to 8

Examples 5 to 8

No Comments

(5) A bag contains  6  red and  4 white balls. Three balls are drawn at random. Find the probability that they are of  the same colour.

Soln. Let, A  :  Event of  getting two balls of  same colour.

P ( A ) = \frac{_{6}C_{3} + _{4}C_{3}}{_{10}C_{3}}  =  \frac{\frac{6 \times 5\times 4}{3 \times 2 \times 1} +\frac{4 \times 3\times 2}{3 \times 2 \times 1}}{\frac{10 \times 9 \times 8}{3 \times 2 \times 1}}  =  \frac{20+4}{120} =   \frac{24}{120}  = \frac{1}{5}

(6) Two dice are thrown together. What is the probability that the sum of  the numbers on the two faces is divisible by  6  or  4 ?

Soln. Here,  n   =   6 \times 6   =  36.

Let, A : Event of  that the sum of  the numbers on the two faces is divisible by 4 or 6

\therefore   A   =   { ( 1, 5 ), ( 2, 4 ), ( 3, 3 ), ( 4, 2 ), ( 5, 1 ), ( 6, 6 ), ( 1, 3 ), ( 2, 2 ), ( 3, 1 ),

( 2, 6 ), ( 3, 5 ), ( 4, 4 ), ( 5, 3 ), ( 6, 2 ) }

m  =  14

\therefore  P (A)  =  \frac{m}{n}  =  \frac{14}{36}  =  \frac{7}{18}.

or

Here,  n   =   6 \times 6   =  36.

Let, A : Event of  that the sum of  the numbers on the two faces is divisible by 4

\therefore A   =   { ( 1, 5 ), ( 2, 4 ), ( 3, 3 ), ( 4, 2 ), ( 5, 1 ), ( 6, 6 ) }  i.e.  m  =  6

\therefore P (A)  =  \frac{m}{n}  = \frac{6}{36}

Let, B : Event of  that the sum of  the numbers on the two faces is divisible by 6

\therefore B   =   { ( 1, 3 ), ( 2, 2 ), ( 3, 1 ), ( 2, 6 ), ( 3, 5 ), ( 4, 4 ), ( 5, 3 ), ( 6, 2 ), ( 6, 6 ) }

i.e.  m  =  9

\therefore  P (B)  =  \frac{m}{n}  = \frac{9}{36}

Let, A \cap B : Event of  that the sum of  the numbers on the two faces is divisible by

12  ( i.e. by  4 and 6 )

\therefore A \cap B   =   { ( 6, 6 ) }            i.e.  m  =  1

\therefore  P (A \cap B)  =  \frac{m}{n}  = \frac{1}{36}

\therefore  P (A \cup B) =     P (A) + P (B) – P (A \cap B)

=    \frac{6}{36}   +  \frac{9}{36}  – \frac{1}{36}

=     \frac{14}{36}    =   \frac{7}{18} .

(7) Two cards are drawn at random from a pack of  52 cards. What is the probability that either both are red or both are kings ?

Soln. Let, A : Both are red  and  B : Both are kings  then  A B : both are kings of  red colour.

Now, by using the formula P (A \cup B)  =  P (A) + P (B) – P (A \cap B)

We have,  P  (A \cup B) = \frac{_{26}\textup{C}_{2} + _{4}\textup{C}_{2} - _{2}\textup{C}_{2}}{_{52}C_{2}}

=  \frac{\frac{26 \times 25}{2!}+ \frac{4 \times 3}{2!}-1}{\frac{52 \times 51}{2!}}

= \frac{325+6-1}{26 \times 51}  =    \frac{330}{26 \times 51}    =   \frac{55}{221}

(8) There are  15 windcheaters in a shop, 3 each of  orange, green, red, blue and pink. Find the probability that Rohit buys a pink windcheater.

Soln. Here,  n  =  _{15}\textup{C}_{1}  =  15

Let, A  :  Rohit buys a pink windcheaters.

\therefore  m  =  _{3}\textup{C}_{1}   =  3

\therefore  P ( A )   =    \frac{m}{n}   =  \frac{3}{15}  =  \frac{1}{5} .

 

Click to next

Share
educational sutra maths sutr Examples mathematics SUMS ganit example concept math proof Parin theorem sir theorems bhai definition Mehta definitions maheta fundamental classes basic class base education sum online dakhla tutorial formula

Leave a Reply