Robert Bosch Probability Quiz 2

Solution:
Prob. (At least 1 black) = 1 – Prob. (None black)
So Prob. (At least 1 black) = 1 – (\frac{^{8}C_{4}}{^{12}C_{4}})
= 1 – \frac{14}{99}
= \frac{85}{99}

Solution:
2 men means other 2 woman and children
So prob. = \frac{^{3}C_{2} \times ^{7}C_{2} }{^{10}C_{4}} = \frac{3}{10}

Solution:
Multiples of 3 up to 120 = \frac{120}{3} = 40
Multiples of 5 up to 120 = \frac{120}{5}= 24 (take only whole number before the decimal part)
Multiple of 15 (3 \times 5) up to 120 = \frac{120}{15} = 8
So total such numbers are = 40 + 24 – 8 = 56
So required probability = \frac{56}{120} = \frac{7}{15}

Solution:
Prob. of 1st winning = \frac{2}{7}, so not winning
= 1 – \frac{2}{7}= \frac{5}{7}
Prob. of 2nd winning = \frac{3}{5}, so not winning
= 1 – \frac{3}{5} =\frac{2}{5}
So required prob. = \frac{2}{7}, * \frac{2}{5} + \frac{3}{5} * \frac{5}{7}= \frac{19}{35}

Solution:
There are 52 cards, out of which there are 12 face cards.
So probability of 2 face cards = \frac{^{12}C_{2}}{^{52}C_{2}} = \frac{11}{221}

Solution:
Case 1: 1 boy and 4 girls
Prob. = ^{5}C_{1} \times \frac{^{4}C_{4}}{^{9}C_{5}} =\frac{5}{146}
Case 2: 2 boys and 3 girls
Prob. = ^{5}C_{2} \times \frac{^{4}C_{3}}{^{9}C_{5}} = \frac{40}{126}
Add the two cases = \frac{45}{126} = \frac{5}{12}

Solution:
Three cases
Case 1: one red, 1 blue
Prob = ^{2}C_{1} \times \frac{^{4}C_{1}}{^{12}C_{2}} =\frac{4}{33}
Case 2: one red, 1 white
Prob = ^{2}C_{1} \times \frac{^{6}C_{1}}{^{12}C_{2}} =\frac{2}{11}
Case 3: one white, 1 blue
Prob = ^{6}C_{1} \times \frac{^{4}C_{1}}{^{12}C_{2}} =\frac{4}{11}
Add all cases = \frac{4}{33} + \frac{2}{11}+ \frac{4}{11}= \frac{2}{3}

Solution:
Probability of choosing one blue ball = 1 –(\frac{4}{15}+ \frac{2}{5}) = \frac{1}{3}

Solution:
Probability of choosing one blue ball is \frac{1}{3}.
And total blue balls are 10. So with \frac{10}{30} we get probability as \frac{1}{3}
So total balls must be 30

Solution:
There are 30 balls in the urn.
Multiples of 2 up to 30 = \frac{30}{2}= 15
Multiples of 3 up to 30 = \frac{30}{3}= 10 (take only whole number before the decimal part)
Multiples of 6 (2\times3) up to 30 = \frac{30}{6} = 5
So total such numbers are = 15 + 10 – 5 = 20
So required probability = \frac{20}{30} = \frac{2}{3}