Permutation and Combination -05

Explanation:-

$$\eqalign{ & {\text{Required}}\,{\text{Money}} \cr & {\text{ = }}\,{\text{P}}{\text{.W}}{\text{.}}\,{\text{of}}\,{\text{Rs}}{\text{.}}\,{\text{10028}}\,{\text{due}}\,{\text{9}}\,{\text{months}}\,{\text{hence}} \cr & = Rs.\,\left[ {\frac{{10028 \times 100}}{{100 + \left( {12 \times \frac{9}{{12}}} \right)}}} \right] \cr & = Rs.\,9200. \cr}$$

Explanation:-

$$\eqalign{ & {\text{P}}{\text{.W}}{\text{.}} = Rs.\left( {2562 - 122} \right) \cr & = Rs.2440 \cr & \therefore {\text{S}}{\text{.I}}{\text{.}}\,{\text{on}}\,{\text{Rs}}{\text{.}}\,{\text{2440}}\,{\text{for}}\,{\text{4}}\,{\text{months}}\,{\text{is}}\,{\text{Rs}}{\text{.}}\,{\text{122}}{\text{.}} \cr & \therefore {\text{Rate}} = \left[ {\frac{{100 \times 122}}{{2440 \times \frac{1}{3}}}} \right]\% = 15\% \cr}$$

Explanation:-

$$\eqalign{ & {\text{C}}{\text{.P}}{\text{.}} = Rs.\,3000. \cr & {\text{S}}{\text{.P}}{\text{.}} = Rs.\,\left[ {\frac{{3600 \times 100}}{{100 + \left( {10 \times 2} \right)}}} \right] \cr & = Rs.\,3000. \cr & {\text{Gain}} = 0\% . \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{number}}\,{\text{of}}\,{\text{balls}} \cr & = \left( {6 + 8} \right) \cr & = 14. \cr & {\text{Number}}\,{\text{of}}\,{\text{white}}\,{\text{balls}} = 8. \cr & P\left( {{\text{drawing}}\,{\text{a}}\,{\text{whiteall}}} \right) \cr & = \frac{8}{{14}} = \frac{4}{7} \cr}$$

Explanation:-

$$\eqalign{ & {\text{Let}}\,{\text{S}}\,{\text{be}}\,{\text{the}}\,{\text{sample}}\,{\text{space}}{\text{.}} \cr & {\text{Then}},\,n\left( S \right) = {}^{52}{C_2} = \frac{{\left( {52 \times 51} \right)}}{{\left( {2 \times 1} \right)}} = 1326. \cr & {\text{Let}}\,{\text{E}}\,{\text{ = }}\,{\text{event}}\,{\text{of}}\,{\text{getting}}\,{\text{1}}\,{\text{spade}}\,{\text{and}}\,{\text{1}}\,{\text{heart}}{\text{.}} \cr & \therefore n\left( E \right) = {\text{number}}\,{\text{of}}\,{\text{ways}}\,{\text{of}}\,{\text{choosing}}\,{\text{1}}\,{\text{spade}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{out}}\,{\text{of}}\,{\text{13}}\,{\text{and}}\,{\text{1}}\,{\text{heart}}\,{\text{out}}\,{\text{of}}\,{\text{13}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {{}^{13}{C_1} \times {}^{13}{C_1}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {13 \times 13} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 169. \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{169}}{{1326}} = \frac{{13}}{{102}}. \cr}$$

Explanation:-

Let S be the sample space.
Then, n(S) = number of ways of drawing 3 balls out of 15 $$\eqalign{ & = {}^{15}{C_3} \cr & = \frac{{\left( {15 \times 14 \times 13} \right)}}{{\left( {3 \times 2 \times 1} \right)}} \cr & = 455. \cr & {\text{Let}}\,{\text{E}}\,{\text{ = }}\,{\text{event}}\,{\text{of}}\,{\text{getting}}\,{\text{all}}\,{\text{the}}\,{\text{3}}\,{\text{red}}\,{\text{balls}} \cr & \therefore n\left( E \right) = {}^5{C_3} = {}^5{C_2} \cr & = \frac{{\left( {5 \times 4} \right)}}{{\left( {2 \times 1} \right)}} = 10. \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{{10}}{{455}} \cr & = \frac{2}{{91}}. \cr}$$

Explanation:-

Here, n(S) = 52.
Let E = event of getting a queen of club or a king of heart.
Then, n(E) = 2. $$\eqalign{ & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{2}{{52}} \cr & = \frac{1}{{26}} \cr}$$

Explanation:-

Clearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.Then
E = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) } $$\eqalign{ & \therefore n\left( E \right) = 15. \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{{15}}{{36}} \cr & = \frac{5}{{12}} \cr}$$

Explanation:-

$$\eqalign{ & P\left( {{\text{getting}}\,{\text{a}}\,{\text{prize}}} \right) \cr & = \frac{{10}}{{10 + 25}} \cr & = \frac{{10}}{{35}} \cr & = \frac{2}{7} \cr}$$

Explanation:-

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then, n(S) = Number ways of selecting 3 students out of 25. $$\eqalign{ & = {}^{25}{C_3} \cr & = \frac{{\left( {25 \times 24 \times 23} \right)}}{{\left( {3 \times 2 \times 1} \right)}} \cr & = 2300. \cr & n\left( E \right) = \left( {{}^{10}{C_1} \times {}^{15}{C_2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {10 \times \frac{{\left( {15 \times 14} \right)}}{{\left( {2 \times 1} \right)}}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1050. \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}}. \cr}$$