Robert Bosch Probability Quiz 1

Solution:
There will be 2 cases
Case 1: 2 red, 1 blue or green
Prob. = ^{6}C_{2} ×\frac{ ^{6}C_{1}}{ ^{12}C_{3}} = \frac{9}{22}
Case 2: all 3 red
Prob. = \frac{^{6}C_{3}}{^{12}C_{3}} = \frac{2}{22}
Add the cases, required prob. = \frac{9}{22} + \frac{2}{22}
= \frac{11}{22} = \frac{1}{2}

Solution:
Multiples of 4 up to 120 = \frac{250}{4} = 62
Multiples of 7 up to 120 = \frac{250}{7}= 35 (take only whole number before the decimal part)
Multiple of 28 (4\times7) up to 250 = \frac{250}{28} = 8
So total such numbers are = 62 + 35 – 8 = 89
So required probability =\frac{89}{250}

Solution:
Total words that can be formed is 5! = 120
Now vowels together:
Take: S, P, T and AE
So their arrangement is 4! * 2! = 48
So required probability = \frac{48}{120} = \frac{2}{5}

Solution:
Two cases:
Case 1: 0 at last place
So 4 choices for 1st digit, 3 for 2nd, 2 for 3rd and 1 for 4th. So numbers = 4*3*2*1 = 24
Case 2: 6 at last place
For 5-digit number 0 cannot be placed at 1st place or cannot be 1st digit
So 3 choices (1, 3, 5) for 1st digit, 3 for 2nd, 2 for 3rd and 1 for 4th.
So numbers = 3*3*2*1 = 18
So total choices = 24+18 = 42
Number total 5-digit numbers that can be formed from 0, 1, 3, 5, and 6
0 not allowed at 1st place, so 4 choices for 1st place, 4 for 2nd, 3 for 3rd, 2 for 4th and
1 for 5th. Sp total = 4*4*3*2*1 = 96
So required probability = \frac{42}{96} =\frac{7}{16}

Directions (14-15): Steve coogan has brought 3 bags which contains 3 colored balls – Red, Green and Yellow.

Bag 1 contains:
24 green balls. Red balls are 4 more than blue balls. Probability of selecting 1 red ball is 4/13

Bag 2 contains:
Total balls are 8 more than 7/13 of balls in bag 1. Probability of selecting 1 red ball is 1/3. The
ratio of green balls to blue balls is 1 : 2

Bag 3 contains:
Red balls equal total number of green and blue balls in bag 2. Green balls equal total number of
green and red balls in bag 2. Probability of selecting 1 blue ball is 3/14.

Solution:
blue balls in bag 3 are 12
Let x green balls are transferred. So
\frac{12}{(56 + x)} = \frac{3}{16} [56 was number of bags in bag 3 before transfer]
Solve, x = 8
So remaining number of balls in bag 1 = 52-8 = 44

Solution:
4x and x = 5x green balls
4+8 = 12 red balls
So \frac{5x}{(5x+12)} = \frac{5}{11}
Solve, x = 2
5*2 = 10 green balls

Solution:
In any case B speaks truth. Now at most 2 people speak truth for 1 question
So case 1: B and A speaks truth
Probability = \frac{3}{7} * \frac{3}{10} * (1-\frac{5}{6}) = \frac{3}{140}
Case 2: B and C speaks truth
Probability = \frac{3}{7} * ( 1- \frac{3}{10}) * \frac{5}{6} = \frac{5}{20}
Case 3: Only B speaks truth
Probability = \frac{3}{7} * ( 1- \frac{3}{10}3/10) * (1- \frac{5}{6}) = \frac{1}{20}
Add the three cases = \frac{6}{20} + \frac{3}{140} =<br /> \frac{45}{140} = \frac{9}{28}

Solution:
Case 1: Birbal does not speak truth, Akbar speaks truth
So Akbar speaks truth here
Probability that Chandni does not speak truth = \frac{3}{10} * (1 – \frac{3}{7}) * ( 1- \frac{5}{6}) = \frac{1}{35}
Case 2: Birbal speaks truth
So Akbar does not speak truth here
Probability that Chandni does not speak truth = ( 1- \frac{3}{10}) *\frac{3}{7} * ( 1- \frac{5}{6}) = \frac{1}{20}
So total = \frac{1}{35} + \frac{1}{20} = \frac{11}{140}

Solution:
There will be 4 cases
Case 1: even, even, odd
Prob. = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}
Case 2: even, odd, even
Prob. = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}
Case 3: odd, even, even
Prob. = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}
Case 4: odd, odd, odd
Prob. = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}
Add all the cases, required prob. = \frac{1}{2}

Solution:
Case 1:first green, second red
Prob. = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72}
Case 2:first red, second green
Prob. = \frac{5}{9} \times \frac{3}{8} = \frac{15}{72}
Add the two cases
\frac{20}{72} + \frac{15}{72} = \frac{35}{72}