TCS NQT Probability Questions Quiz 1

Non defective blue ball + Non defective pink ball =146
(5/9)(2x/3)+(7/8)(1x/3)=146
10x/27+7x/24=146
x=432

Total number of student given are 100.

Number of student passed in first examination is 60. So, probability of student passing in first examination is, P(A) = = 0.6

Number of student passed in second examination is 50. So, probability of student passing in second examination is, P(B) = = 0.5

Number of student passed in both examination is 30. So, probability of student passing in both examination is, P(A ∩ B) = = 0.3

From probability addition rule,

P(A ∪ B) = P(A) + P(B) - P(A∩B)

P(A ∪ B) = 0.6 + 0.5 - 0.3

P(A ∪ B) = 0.8

Thus, the probability of students passing at least in one subject is 0.8.

The probability of student failing in both subject is given by,

P(A U B)'=1 - P(A ∪ B)

P(A U B)'= 1- 0.8

P(A U B)'=0.2

Thus, Probability of student failing in both subjects are 0.2.



 

From a bag containing 8 green and 5 red balls.
Therefore total balls are 8+5=13

three drawn one after the other.

the probability of all three balls beings green if the balls drawn are replace before the next ball pick = (8/13)(8/13)(8/13)
= 512/2197

the probability of all three balls drawn are not replaced (8/13)(7/12)(6/11)
= 336/1716
= 28/143

Bag 1:

White balls = 8

Blue balls = 3

Total balls = 8+3=11

Bag 2:

White balls = 7

Blue balls = 4

Total balls = 7+4=11

Total blue balls = 3+4=7

Total balls in both bags = 11+11=22

So, Probability of getting blue ball = Number of blue balls / Total number of balls = 7/22

P(Getting a red ball from the first bag): (1/2)*(5/12)
P(Getting a red ball from the 2nd bag): (1/2)*(3/15)
P(Drawing a red ball):
(1/2)*(5/12)+(1/2)*(3/15)= 37/120 (Ans).

total events = 6*6*6=216
possible cases for sum equal to 10 are
(1,3,6)-6 combinations (3!)
(1,4,5)-6 combinations
(2,3,5)-6 combinations
(2,4,4)-3 combinations (3!/2! as repetition)
(3,3,4)-3 combinations
(2,2,6)-3 combinations
so total combinations are 27
so probability will be 27/216

There are 5 letters and 5 addressed envelopes. The formula to find the number derangement is given by :-



For n= 5 ,

D_{5}= 5!(1-\frac{1}{1!}+\frac{2}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!})

=(120)\times \frac{11}{30}=44

we have Number of ways that all the letters may be placed in wrongly addressed envelopes is 44. Total number of ways to put 5 letters in 5 envelope = 5!=120 Then, the probability that all the letters may be placed in wrongly addressed envelopes is

=44/120= 11/30

Answer: c
Probability of thangam being selected= 1/3 and him not being selected = 1-1/3=2/3
Probability of pandiyamma being selected=1/5 and him not being selected= 1-1/5=4/5
Probability that both are not selected = 2/3*4/5=8/15

no. of numbers between 1 to 999 in which at-least one 2 is present =
[ fix 2 at 100th position and the unit and 10th places can be filled in 10*10 ways (as each place can be filled with 0 to 9 ) ]
+ [ fix 2 at 10th position and then 100th place can be filled with 9 values (0 to 9 excluding 2, as we have already been counted 2 at the 100th place) * 10 ( as unit place can still use values 0 to 9 ) ]
+ [ fix 2 at unit place and then 100th place can be filled with 9 values (0 to 9 excluding 2, as we have already been counted 2 at 100th place) * 10th place can be filled with 9 values (0 to 9 excluding 2, as we have already been counted 2 at 10th place) ]
=10*10+9*10+9*9
=100 + 90 + 81
=271
Now, no. of numbers between 1000 to 1099 in which at-least one 2 is present = 2 is fixed at 10th place and 10 ways of filling unit place
+ 9 ways of filling 10th place (0 to 9 excluding 2 ) and 2 is fixed at unit place
=10 + 9
=19

So, total no. of tickets on which digit 2 is appearing = 271 + 19 = 290
Therefore, required probability = 290/1100