Probability — Part 1 MCQs with Explanations

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

Total number of balls = (2 + 3 + 2) = 7 Let S be the sample space Then, n(S) = Number of ways of drawing 2 balls out of 7 $$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^7{C_2} \cr & {\text{n}}\left( {\text{S}} \right) = \frac{{\left( {7 \times 6} \right)}}{{\left( {2 \times 1} \right)}} \cr & {\text{n}}\left( {\text{S}} \right) = 21 \cr} $$ Let E = Event of 2 balls, none of which is blue ∴ n(E) = Number of ways of drawing 2 balls out of (2 + 3) balls $$\eqalign{ & {\text{n}}\left( {\text{E}} \right)\, = {}^5{C_2} \cr & {\text{n}}\left( {\text{E}} \right) = \frac{{\left( {5 \times 4} \right)}}{{\left( {2 \times 1} \right)}} \cr & {\text{n}}\left( {\text{E}} \right) = 10 \cr & \therefore {\text{P}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} = \frac{{10}}{{21}} \cr} $$

Total number of balls = (8 + 7 + 6) = 21 Let E = event that the ball drawn is neither red nor green = event that the ball drawn is blue $$\eqalign{ & \therefore n(E) = 7 \cr & \therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{7}{{21}} = \frac{1}{3} \cr} $$

In two throws of a dice, n(S) = (6 x 6) = 36 Let E = event of getting a sum = {(3, 6), (4, 5), (5, 4), (6, 3)} $$\eqalign{ & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{4}{{36}}\cr & = \frac{1}{9} \cr} $$

Getting at most Two heads means 0 to 2 but not more than 2 Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} Let E = event of getting at most two heads Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT} $$\therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{7}{8}$$

In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36 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), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} $$\eqalign{ & \therefore n\left( E \right) = 27 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{27}}{{36}} = \frac{3}{4} \cr} $$

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 $$\eqalign{ & = {}^{25}{C_3} \cr & = \frac{{ {25 \times 24 \times 23} }}{{ {3 \times 2 \times 1} }} \cr & = 2300 \cr & n\left( E \right) = {^{10}{C_1}{ \times ^{15}}{C_2}} \cr & = {10 \times \frac{{ {15 \times 14} }}{{ {2 \times 1} }}} \cr & = 1050 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}} \cr} $$

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

Let S be the sample space $$\eqalign{ & {\text{Then}},n\left( S \right) = {}^{52}{C_2} \cr & = \frac{{ {52 \times 51} }}{{\left( {2 \times 1} \right)}} \cr & = 1326 \cr} $$ Let E = event of getting 2 kings out of 4 $$\eqalign{ & \therefore n\left( E \right) = {}^4{C_2} = \frac{{ {4 \times 3} }}{{ {2 \times 1} }} = 6 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{6}{{1326}} \cr & = \frac{1}{{221}} \cr} $$

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)} $$\eqalign{ & \therefore n\left( E \right) = 15 \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} \cr & = \frac{{15}}{{36}} \cr & = \frac{5}{{12}} \cr} $$ Alternate solution Understanding the Question: We're tossing two dice. Each die has six sides (1, 2, 3, 4, 5, 6). We want to find the probability that the sum of the numbers shown on both dice is a prime number. A prime number is a number greater than 1 that is only divisible by 1 and itself (e.g., 2, 3, 5, 7, 11...). Finding the Possible Outcomes: First, let's figure out all the possible sums we can get when we add the numbers on two dice. The minimum sum is 2 (1+1) and the maximum is 12 (6+6). Identifying Prime Number Sums: Now, let's list the sums that are prime numbers: 2, 3, 5, 7, 11. Counting Favorable Outcomes: Let's count how many ways we can get each of these prime sums: * 2: Only one way (1+1) * 3: Two ways (1+2, 2+1) * 5: Four ways (1+4, 2+3, 3+2, 4+1) * 7: Six ways (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) * 11: Two ways (5+6, 6+5) Total favorable outcomes (prime number sums): 1 + 2 + 4 + 6 + 2 = 15 Calculating Total Possible Outcomes: The total number of possible outcomes when tossing two dice is 6 (outcomes for the first die) * 6 (outcomes for the second die) = 36 Calculating Probability: Probability is calculated as (Favorable Outcomes) / (Total Possible Outcomes). So, the probability of getting a prime number sum is 15/36. This simplifies to 5/12. Therefore, the correct option is B: 5/12

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

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

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

Clearly, there are 52 cards, out of which there are 12 face cards. $$\eqalign{ & \therefore P\left( {{\text{getting}}\,{\text{a}}\,{\text{face}}\,{\text{card}}} \right) \cr & = \frac{{12}}{{52}} = \frac{3}{{13}} \cr} $$

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

Here, S = {1, 2, 3, 4,........, 19, 20} Let E = even of getting a multiple of 3 = {3, 6, 9, 12, 15, 18} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{20}} = \frac{3}{{10}}$$

Total number of marbles = (6 + 4 + 2 + 3) = 15 Let E be the event of drawing 4 marbles such that none is blue. Then, n(E) = number of ways of drawing 4 marbles out of 11 non-blue. $${}^{11}\mathop C\nolimits_4 = \frac{{11 \times 10 \times 9 \times 8}}{{4 \times 3 \times 2 \times 1}}$$ = 330 And n(S) = $${}^{15}\mathop C\nolimits_4 = $$ $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$ = 1365 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}}$$ $$ = \frac{{330}}{{1365}}$$ $$ = \frac{{22}}{{91}}$$ ∴ Required probality =$$\left( {1 - \frac{{22}}{{91}}} \right)$$ $$ = \frac{{69}}{{91}}$$

Here S = {HH, HT, TH, TT} Let E = event of getting at least one head = {HT, TH, HH} $$\therefore P\left( E \right) = \frac{{n(E)}}{{n(S)}} = \frac{3}{4}$$

Number of Red balls = 4 Number of Blue balls = 6 Number of Pink balls = 8 Total number of balls = 4 + 6 + 8 = 18 Required probability $$\eqalign{ & = \frac{4}{{18}} \times \frac{{11}}{{20}} + \frac{6}{{18}} \times \frac{{11}}{{20}} \cr & = \frac{{11}}{{20}}\left[ {\frac{4}{{18}} + \frac{6}{{18}}} \right] \cr & = \frac{{11}}{{20}} \times \frac{{10}}{{18}} \cr & = \frac{{11}}{{36}} \cr} $$

Number of blue balls = 3 Number of green balls = 2 Numbers of red balls = 5 Total balls in the bag = 3 + 2 + 5 = 10 Total possible outcomes = Selection of 4 balls out of 10 balls $$ = {}^{10}\mathop C\nolimits_4 = \frac{{10!}}{{4! \times (10 - 4)!}}$$ $$ = \frac{{10 \times 9 \times 8 \times 7}}{{1 \times 2 \times 3 \times 4}}$$ = 210 Favorable outcomes = (selection of 2 green balls out of 2 balls) × (selection of 2 balls out of 3 blue balls) $$\eqalign{ & = {}^2\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_2 \cr & = 1 \times 3 \cr & = 3 \cr} $$ ∴ Required probability $$ = \frac{{{\text{Favorable outcomes}}}}{{{\text{Total possible outcomes}}}}$$ $$\eqalign{ & = \frac{3}{{210}} \cr & = \frac{1}{{70}} \cr} $$

Total number of balls = (6 + 2 + 4 + 3) = 15 Let E be the event of drawing 5 balls out of 9 non-blue balls. $$ = {}^9\mathop C\nolimits_5 = {}^9\mathop C\nolimits_{\left( {9 - 5} \right)} = {}^9\mathop C\nolimits_4 $$ $$ = \frac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2 \times 1}}$$ = 126 And, n(S) = $${}^{15}\mathop C\nolimits_5 = $$ $$\frac{{15 \times 14 \times 13 \times 12 \times 11}}{{5 \times 4 \times 3 \times 2 \times 1}}$$ = 3003 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = $$ $$\frac{{126}}{{3003}} = \frac{6}{{143}}$$ ∴ Required Probability $$\eqalign{ & = \left( {1 - \frac{6}{{143}}} \right) \cr & = \frac{{137}}{{143}} \cr} $$

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} Let E = event of getting at least two heads = {THH, HTH, HHT, HHH} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{8} = \frac{1}{2}$$

In a simultaneous throw of two dice, we have n (S) = (6 × 6) = 36 Let E = event of getting a total of 10 or 11 = [(4, 6), (5, 5), (6, 4), (5, 6), (6, 5)] $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{5}{{36}}$$

Hence, n (S) = 52 There are 13 spades (including one ten) and there are 3 more ten Let E = event of getting a ten or a spade Then, n (E) = (13 + 3) = 16 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{16}}{{52}} = \frac{4}{{13}}$$

Total number of balls = (4 + 5 + 6) = 15 P (drawing a red ball or a green ball) = P (red) + P (green) $$\eqalign{ & = \left( {\frac{4}{{15}} + \frac{5}{{15}}} \right) \cr & = \frac{9}{{15}} \cr & = \frac{3}{5} \cr} $$

Number of red balls = 6 Number of yellow balls = 11 Number of pink balls = 5 Total number of balls = 6 + 11 + 5 = 22 Total possible outcomes $$n(E) = {}^{22}\mathop C\nolimits_2 = \frac{{22!}}{{2!(22 - 2)!}}$$ $$ = \frac{{22!}}{{2! \times 20!}} $$ $$ = \frac{{22 \times 21}}{{2 \times 1}} $$ $$=$$ 231 Number of favourable outcomes $$n(S) = {}^6\mathop C\nolimits_1 \times {}^{11}\mathop C\nolimits_1 $$ = 6 × 11 = 66 Required probability = $$\frac{{n(E)}}{{n(S)}}$$ $$ = \frac{{66}}{{231}}$$ $$ = \frac{1}{7}$$

Let $${E_1}$$ = event that the husband is selected and $${E_2}$$ = event that the wife is selected. Then, $$P({E_1}) = \frac{1}{7}{\text{ }}$$ and $${\text{ }}P({E_2}) = \frac{1}{5}$$ ∴ $$P({\overline E _1}) = \left( {1 - \frac{1}{7}} \right)$$ $${\text{ = }}\frac{6}{7}$$ and $$P({\overline E _2}) = $$ $$\left( {1 - \frac{1}{5}} \right) = \frac{4}{5}$$ ∴ Required probability = P [( A and not B) or (B and not A)] = P $$\left[ {\left( {{E_1} \cap {{\bar E}_2}} \right)\,{\text{or}}\,\left( {{E_2} \cap {{\bar E}_1}} \right)} \right]$$ = P $$\left( {{E_1} \cap {{\overline E }_2}} \right) + {\text{ P}}\left( {{E_2} \cap {{\overline E }_1}} \right)$$ = $${\text{P}}\left( {{E_1}} \right).{\text{ P}}\left( {{{\overline E }_2}} \right) + P\left( {{E_2}} \right).P\left( {{{\overline E }_1}} \right)$$ $$\eqalign{ & = \left( {\frac{1}{7} \times \frac{4}{5}} \right) + \left( {\frac{1}{5} \times \frac{6}{7}} \right) \cr & = \frac{10}{{35}} \cr & = \frac{2}{7} \cr} $$

We know that in a simultaneous throw of two dice, n(S) = 6 × 6 = 36 Let E = event of getting a total of 7 = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{36}} = \frac{1}{6}$$

Here, n(S) = 52 There are 26 red cards (including 2 kings) and there are 2 more kings. Let E = event of getting a red card or a king. Then, n(E) = 28 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{28}}{{52}} = \frac{7}{{13}}$$

Clearly, n (S) = $$n{\text{ }}(S) = $$ $${}^{52}\mathop C\nolimits_2 = $$ $$\frac{{\left( {52 \times 51} \right)}}{2}$$ = 1326 Let $${{E_1}}$$ = event of getting both red cards $${{E_2}}$$ = event of getting both kings Then, $${{E_1}}$$ $$ \cap $$ $${{E_2}}$$ = event of getting 2 kings of red cards. ∴ $$n{\text{ }}({E_1}) = {}^{26}\mathop C\nolimits_2 = \frac{{\left( {26 \times 25} \right)}}{{\left( {2 \times 1} \right)}}$$ = 325 and $$n{\text{ }}({E_2}) = {}^4\mathop C\nolimits_2 = \frac{{\left( {4 \times 3} \right)}}{{\left( {2 \times 1} \right)}}$$ = 6 $$n\left( {{E_1} \cap {E_2}} \right) = {}^2{C_2} = 1$$ $$\therefore P({E_1}) = \frac{{n({E_1})}}{{n(S)}} = \frac{{325}}{{1326}}$$ and $$P({E_2}) = \frac{{n({E_2})}}{{n(S)}} = \frac{6}{{1326}}$$ $$P({E_1} \cap {E_2}) = \frac{1}{{1326}}$$ ∴ P (both red or both kings) $$\eqalign{ & = P\left( {{E_1} \cup {E_2}} \right) \cr & = P\left( {{E_1}} \right) + P\left( {{E_2}} \right) - P\left( {{E_1} \cap {E_2}} \right) \cr & = \left( {\frac{{325}}{{1326}} + \frac{6}{{1326}} - \frac{1}{{1326}}} \right) \cr & = \frac{{330}}{{1326}} \cr & = \frac{{55}}{{221}} \cr} $$

Total number of balls = (6 + 4 + 2 + 3) = 15 Let E be the event of drawing 2 red balls. Then, n(E) $$ = {}^6\mathop C\nolimits_2 $$ $$ = \frac{{6 \times 5}}{{2 \times 1}}$$ = 15 Also, $$n(S) = {}^{15}\mathop C\nolimits_2 $$ $$ = \frac{{15 \times 14}}{{2 \times 1}}$$ = 105 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{15}}{{105}} = \frac{1}{7}$$

Total number of marbles = (4 + 5 + 3) = 12 Let E be the event of drawing 3 marbles such that none is blue. Then, n (E) = number of ways of drawing 3 marbles out of 7 = $${}^7\mathop C\nolimits_3 $$ $$ = \frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}}$$ = 35 And, $$n(S) = {}^{12}\mathop C\nolimits_3 $$ $$ = \frac{{12 \times 11 \times 10}}{{3 \times 2 \times 1}}$$ = 220 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{35}}{{220}} = \frac{7}{{44}}$$ ∴ Required probability = 1 - P(E) = $$\left( {1 - \frac{7}{{44}}} \right)$$ = $$ \frac{{37}}{{44}}$$

P (none is defective) = n (E) = $$\frac{{{}^{16}\mathop C\nolimits_2 }}{{{}^{20}\mathop C\nolimits_2 }} = $$ $$\left( {\frac{{16 \times 15}}{{2 \times 1}} \times \frac{{2 \times 1}}{{20 \times 19}}} \right)$$ $$ = \frac{{12}}{{19}}$$ P (at least 1 is defective) =$$\left( {1 - \frac{{12}}{{19}}} \right)$$ $$ = \frac{7}{{19}}$$

Total number of marbles = (6 + 4 + 2 + 3) = 15 Let E be the event of drawing 2 marbles such that either both are green or both are yellow. Then, n (E) = $$\left( {{}^2\mathop C\nolimits_1 + {}^3\mathop C\nolimits_2 } \right)$$ $$ = \left( {1 + {}^3\mathop C\nolimits_1 } \right)$$ = (1 + 3) = 4 And, n (S) = $${}^{15}\mathop C\nolimits_2 = $$ $$\frac{{15 \times 14}}{{2 \times 1}}$$ = 105 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{105}}$$

When a die is thrown, we have S = {1, 2, 3, 4, 5, 6} Let, E = event of getting a number greater than 4 = {5, 6} $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{2}{6} = \frac{1}{3}$$

Clearly, there are 52 cards, out of which there are 12 face cards 4 jack, 4 queens, and 4 kings ∴ P (getting a face card) $$ = \frac{{12}}{{52}}$$ $$ = \frac{3}{{13}}$$

Total number of marbles = (6 + 4 + 2 + 3) = 15 Let E be the event of drawing 1 green, 2 blue and 1 red marble. Then, n (E) = $${}^2\mathop C\nolimits_1 \times {}^4\mathop C\nolimits_2 \times {}^6\mathop C\nolimits_1 $$ $$ = 2 \times \frac{{4 \times 3}}{{2 \times 1}} \times 6$$ = 72 And, n (S) = $${}^{15}\mathop C\nolimits_4 = $$ $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$ = 1365 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{72}}{{1365}}$$ $$ = \frac{{24}}{{455}}$$

Total number of balls = (2 + 3 + 2) = 7 Let, E be the event of drawing 2 non-blue balls. Then, n (E) = $${}^5\mathop C\nolimits_4 = \frac{{5 \times 4}}{{2 \times 1}}$$ = 10 And, n (S) = $${}^7\mathop C\nolimits_2 = \frac{{7 \times 6}}{{2 \times 1}}$$ = 21 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{10}}{{21}}$$

Total number of balls = (10 + 10) = 20 Let E be the event of drawing 2 balls of the same colour. n (E) = number of ways of drawing 2 black balls or 2 white balls. n (E) = $$\left( {{}^{10}\mathop C\nolimits_2 \times {}^{10}\mathop C\nolimits_2 } \right)$$ $$ = 2 \times {}^{10}\mathop C\nolimits_2 $$ $$ = 2 \times \frac{{10 \times 9}}{{2 \times 1}}$$ = 90 n (S) = number of ways of drawing 2 balls out of 20 balls $$ = {}^{20}\mathop C\nolimits_2 = \frac{{20 \times 19}}{{2 \times 1}}$$ = 190 $$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{90}}{{190}} = \frac{9}{{19}}$$

Out of mangoes, 4 mangoes are rotten ∴ Required probability $$\eqalign{ & = \frac{{{}^6{C_2}}}{{{}^{10}{C_2}}} \cr & = \frac{{\frac{{6!}}{{2!\left( {6 - 2} \right)!}}}}{{\frac{{10!}}{{2!\left( {10 - 2} \right)!}}}} \cr & = \frac{{\frac{{6!}}{{2!4!}}}}{{\frac{{10!}}{{2! \times 8!}}}} \cr & = \frac{{\frac{{6 \times 5}}{{1 \times 2}}}}{{\frac{{10 \times 9}}{{1 \times 2}}}} \cr & = \frac{{6 \times 5}}{{10 \times 9}} \cr & = \frac{1}{3} \cr} $$