Robert Bosch Permutation and Combinations Quiz 4

Blue marble – x
\frac{^xC_{1}}{^27C_{1} = \frac{1}{3}
\frac{x}{27}=\frac{1}{3} —> x= \frac{27}{3}=9
No of green marbles = Total – Blue marble =27-9=18

Total Balls =10
Other than red ball = ^6C_{2}
\frac{^6C_{2}}{^10C_{2}} = \frac{1}{3} —> 1-\frac{1}{3} = \frac{2}{3}

Total balls in A bag = 14, Total balls in A bag = 12
A bag = \frac{1}{2}*(\frac{^8C_{1}}{^14C_{1}}) = \frac{2}{7}
B bag = \frac{1}{2}*(\frac{^6C_{1}}{^12C_{1}}) = \frac{1}{4}
—> total ProbaRamesh pokhriyal attended a examination in which there were three sections namely Reasoning, Maths and English. Reasoning part contains 4 questions. There are 5 questions in maths section and 6 questions in English section. If three questions are selected randomly from the list of questions then what is the probability that all of them are from maths? bility = \frac{2}{7}+ \frac{1}{4}
= \frac{15}{28}

Total Marbles = 12
Either all are green or all are red = ^5C_{3} + ^3C_{3}
probability = ^5C_{3} + \frac{^3C_{3}}{^12C_{3}} = \frac{11}{220}= \frac{1}{20}

Total Marbles = 12
other than blue \frac{^8C_{3}}{^12C_{3}}= \frac{14}{55}
probability = 1-\frac{14}{55} = \frac{41}{55}

Total Marbles = 12
Probability = \frac{^5C_{2}}{^12C_{2}} =\frac{5}{33}

 

Total caps = 14
Probability = \frac{^5C_{2}*^2C_{1}}{^14C_{3}} = \frac{5}{91}

Total caps = 14
Probability = \frac{^5C_{2}*^4C_{1}*^2C_{1}}{^14C_{4}} = \frac{80}{1001}

Total caps = 14
Probability = \frac{^5C_{0}*^9C_{3}}{^14C_{3}} = \frac{3}{13}

Balls are picked in two manners – RWRW or WRWR
So probability = (\frac{5}{12})*(\frac{7}{11})*(\frac{4}{10})*(\frac{6}{9}) + (\frac{7}{12})*(\frac{5}{11})*(\frac{6}{10})*(\frac{4}{9}) = \frac{14}{99}