Robert Bosch Probability Quiz 3

Solution:
Only one will pass means the other will fail
Probability that A will pass in exam is \frac{3}{5}.
So Probability that A will fail in exam is 1 – \frac{3}{5} = \frac{2}{5}
Probability that B will pass in exam is \frac{5}{8}.
So Probability that B will fail in exam is 1 – \frac{5}{8} =\frac{3}{8}
So required probability = P(A will pass)*P(B will fail) + P(B will pass)*P(A will fail)
So probability that only one will pass in the exam = \frac{3}{5} * \frac{3}{8} + \frac{5}{8}* \frac{2}{5} = \frac{19}{40}

Solution:
Total events will be 6*6*6 = 216
Favorable events for having same number
is {1,1,1}, {2,2,2}, {3,3,3}, {4,4,4},
{5,5,5}, {6,6,6} – so 6 events
Probability of same number on all dices is \frac{6}{126} = \frac{1}{36}

Solution:
Multiples of 3 up to 150 = \frac{150}{3}= 50
Multiples of 7 up to 150 = \frac{150}{7} = 21 (take only whole number before the decimal part)
Multiples of 21 (3 \times 7) up to 150 = \frac{150}{21} = 7
So total such numbers are = 50 + 21 – 7 = 64
So required probability = \frac{64}{150} = \frac{32}{75}

Solution:
Prime numbers up to 55 is 16 numbers which are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
31, 37, 43, 41, 47, 53.
So probability = \frac{16}{55}

Solution:
Let 2 events A and B
Odds against A are 1 : 3
So probability of occurrence of A = \frac{3}{(1+3)} = \frac{3}{4}.
And non-occurrence of A = \frac{1}{4}
Odds in favor of B are 2 : 5
So probability of occurrence of B = \frac{2}{(2+5)} = \frac{2}{7}.
And non-occurrence of B = \frac{5}{7}
Case 1: A occurs and B does not occur
So probability = \frac{3}{4}* \frac{5}{7} = \frac{15}{28}
Case 2: B occurs and A does not occur
So probability =\frac{2}{7} * \frac{1}{4}= \frac{2}{28}
So probability that one will occur = \frac{15}{28} + \frac{2}{28} =\frac{17}{28}

Solution:
In 52 cards, there are 13 diamond cards and 4 queens.
1 card is chosen at random
For 1 diamond card, probability = \frac{13}{52}
For 1 queen, probability = \frac{4}{52}
For cards which are both diamond and queen, probability = \frac{1}{52}
So required probability = \frac{13}{52} + \frac{4}{52} – \frac{1}{52}= \frac{16}{52}= \frac{4}{13}

Solution:
In 52 cards, there are 26 red cards and 4 ace and there 2 such cards which are both red and ace.
1 card is chosen at random
For 1 red card, probability = \frac{26}{52}
For 1 ace, probability = \frac{4}{52}
For cards which are both red and ace, probability = \frac{2}{52}
So required probability = \frac{26}{52}+\frac{4}{52} – \frac{2}{52} = \frac{28}{52}= \frac{7}{13}

Solution:
Multiples of 3 up to 250 = \frac{250}{3} = 83 (take only whole number before the decimal part)
Multiples of 8 up to 250 = \frac{250}{3} = 31
Multiples of 24 (3 \times 8) up to 250 = \frac{250}{24} = 10
So total such numbers are = 83 + 31 – 10 = 104
So required probability = \frac{104}{250} = \frac{52}{125}

Solution:
Both balls being non-blue means both balls are either white or green
There are total 12 balls (4+3+5)and total 7 white + green balls.
So required probability = \frac{^{7}C_{2}}{^{12}C_{2}}
= [\frac{\frac{(7*6)}{(2*1)} }{\frac{(12*11)}{(2*1)} }]
= \frac{21}{66}= \frac{7}{22}

Solution:
There are total 4+3+5 = 12 balls
Probability of first ball being green is = \frac{5}{12}
Now total green balls in box = 5 – 1 = 4
So total white + green balls = 4 + 4 = 8
So probability of second ball being white or green is \frac{8}{12} = \frac{2}{3}
So required probability = \frac{5}{12} * \frac{2}{3}= \frac{5}{18}