Blog

Parin Sir > Quantitative Aptitude (Maths) > Probability > Examples 9 to 11 > Examples 9 to 11

Examples 9 to 11

No Comments

(9) A committee of  4  has to be formed out of  7 boys and  3  girls. Find the probability that it includes at least 1 girl.

Soln. Here,  n  = _{10}\textup{C}_{4}

Let  A  :  at least 1 girl in committee.

\therefore P ( A ) =  \frac{_{3}\textup{C}_{1} \times _{7}\textup{C}_{3} + _{3}\textup{C}_{2} \times _{7}\textup{C}_{2} + _{3}\textup{C}_{3} \times _{7}\textup{C}_{1}}{_{10}\textup{C}_{4}}

=     \frac{105+63+7}{10\times3\times7}    \left (\because \textup{ }_{10}\textup{C}_{4} = \frac{10\times9\times8\times7}{4 \times 3\times2\times1}=10\times3\times7 \right )

=    \frac{175}{10\times3\times7}  =  \frac{5}{6}

or

Here,  n  = _{10}\textup{C}_{4}

Let  A  :  at least 1 girl in committee.

\therefore A’  :  There is no girl in committee.

\therefore  P (A’ ) =    \frac{_{3}\textup{C}_{0} \times _{7}\textup{C}_{4}}{_{10}\textup{C}_{4}}

=    \frac{7\times6\times5\times 4}{10\times9\times8\times7}    =   \frac{1}{6}

\therefore P (A ) =     1  –  P (A’ )

=      1  – \frac{1}{6}     =  \frac{5}{6}

(10) The odds in favour of an event are  2 : 3. Then find the probability of  occurrence of  the event.

Soln. Here,  A  :  favour of  an event.

\therefore P (A )  = \frac{2}{5}

(11) The odds against of an event are  2 : 3. Then find the probability of  occurrence of  the event.

Soln. Here,  A  :  against of  an event.

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