AMDOCS Probability Problems with Solutions Quiz-1

Probability of 1st ball = red is 5/(4+5+6) = 5/15.

P(2nd ball = red) = 4/14 as only 4 red balls remain if 1st ball = red. Out of 14.

P(3rd ball = red) is 3/13.

Answer = 5/15 * 4/14 * 3/13
= 2/91

Total bulbs = 20
Defective = 4
Non - defective = 20-4 =16

Probability that at least one of these is defective = 4/20
= 1/5

30/100

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p(a/b)=p(a+b)/p(b)

=0.9/0.5

=1.8

Total members = 3+3+5 = 11
Total no. of ways of selection 4 members is 11C4 = 330
No. of ways of selecting exactly 3 client eng managers are
3 from client (AND) 1 from content managers
(OR)
3 from client (AND) 1 from R&D
=> 5C3*3C1 + 5C3*3C1 = 30 + 30 = 60
Prob. ll be 60/330 = 2/11

5/6

Probability of getting 5 = 1/4
Probability of getting 1,2,3,4 or 6 = 1 – 1/4 = 3/4
Therefore
probability of getting 1 = (3/4)/5 = 3/20
probability of getting 2 = (3/4)/5 = 3/20
probability of getting 3 = (3/4)/5 = 3/20
probability of getting 4 = (3/4)/5 = 3/20
probability of getting 6 = (3/4)/5 = 3/20
Event: 3 dice rolled and their sum should be atleast 14 without getting a 6 on any dice.
ie (5,5,5) , (4,5,5),(5,4,5),(5,5,4)
Probability of getting (5,5,5) = 1/4*1/4*1/4 = 1/64
Probability of getting (4,5,5) = 3/20*1/4*1/4 = 3/320
Probability of getting (5,4,5) = 1/4*3/20*1/4 = 3/320
Probability of getting (5,5,4) = 1/4*1/4*3/20 = 3/320
Probability of getting a sum of atlest 14
= 1/64 + 3/320 + 3/320 + 3/320
= 7/160

1/5

Explanation:

no of cases for the last two digits are

66  76 86 96

67  77  87 97

68  78 88 98

69  79  89 99

so there are total 16 cases and no of chances are only 3 so probability of getting access will be 3/16.