Robert Bosch Probability Quiz 4

Solution:
There are 36 total events which can happen
({1,1), {1,2}……………….{6,6})
For a total of 7 on dices, we have – {1,6}, {6,1},
{2,5}, {5,2}, {3,4}, {4,3} – so 6 choices
So required probability = \frac{6}{36}= \frac{1}{6}

Solution:
There are 36 total events which can happen
({1,1), {1,2}……………….{6,6})
For sum of number to be a multiple of 5, we have – {1,4}, {4,1}, {2,3}, {3,2}, {4,6}, {6,4},
{5,5} – so 7 choices
So required probability = \frac{7}{36}

Solution:
There are 5 tickets which contain a multiple of 5
So probability of 1st ticket containing multiple of 5 = \frac{5}{25} = \frac{1}{5}
Now:
Case 1: If the ticket chosen contained 15
If there was a 15 on first draw, then there are 7 tickets in box which contain a multiple of 3 out
of 24 tickets.
( \frac{25}{3}– 1 = 8 – 1 = 7 ) – because 15 is already out from the box
So probability = \frac{7}{24} (24 tickets remaining after 1st draw)
Case 2: If the ticket chosen contained other than
15 (5 or 10 or 20 or 25)
If 15 was not there on first draw, then there are 8 tickets in box which contain a multiple of 3 out
of 24 tickets.
(\frac{25}{3}= 8) – because 15 is already out from the box
So probability = \frac{8}{24} (24 tickets remaining after 1st draw)
Add the cases for probability of multiple of 3 on second ticket, so prob.
= \frac{7}{24}+ \frac{8}{24}= \frac{15}{24}
(added the cases because we want one of these cases to happen and not both)
So required probability = \frac{1}{5} * \frac{15}{24} = \frac{1}{8}
(multiplied the cases because we want both to happen)

Solution:
There are 90 two digit numbers (10-99)
Multiple of 2 = \frac{90}{2} = 45
Multiple of 14 = \frac{90}{14} = 6
Since all multiples of 14 are also multiple of 2,
so favorable events = 45 – 6 = 39
So required probability = \frac{39}{90} = \frac{13}{30}

Solution:
Case 1: first was a white ball
Now it is put in second urn, so total white balls in second urn = 5+1 = 6, and total balls in
second urn = 12+1 = 13
So probability of white ball from second urn = \frac{6}{13}
Case 2: first was a blue ball
Now it is put in second urn, so total white balls in second urn remain 5, and total balls in second
urn = 12+1 = 13
So probability of white ball from second urn = \frac{5}{13}
So required probability = \frac{6}{13} + \frac{5}{13} = \frac{11}{13} (added the cases because we want one of these cases to happen and not both)

Explanation :
Probability = \frac{ (\frac{3}{5} * \frac{5}{8})}{([\frac{3}{5} * \frac{5}{8}] + [\frac{2}{5} * \frac{1}{3}])}
= \frac{45}{61}

Explanation :
P(A) = \frac{^{20}C_{2}}{^{30}C_{2}},
P(B) = \frac{^{22}C_{2}}{^{30}C_{2}}
P(A\cap B ) = \frac{^{15}C_{2}}{^{30}C_{2}}
P(A\cup B ) = P(A) + P(B) – P(A\cap B )
= (\frac{^{20}C_{2}}{^{30}C_{2}})+(\frac{^{22}C_{2}}{^{30}C_{2}})-(\frac{^{15}C_{2}}{^{30}C_{2}})
= \frac{316}{435}

Explanation :
Five persons is to be chosen from a group of 10 people = ^{10}C_{5} 10C5 = 252
Couple Serve together = ^{8}C_{3} * ^{2}C_{2} = 56
Couple does not serve = ^{8}C_{5} = 56
Probability = \frac{102}{252} =\frac{51}{126}

Explanation :
Man only = ^{8}C_{2} = 14
Probability of selecting no woman = \frac{14}{91}
Probability of selecting atleast one woman = 1 –\frac{14}{91} = \frac{77}{91}

Explanation :
Total probability = ^{6}C_{2} = 15
Probability that one is Banana and another one is
orange = ^{3}C_{1} * ^{3}C_{1} = 9
probability =\frac{9}{15} = \frac{3}{5}