TCS Probability Questions with Answers Quiz-1

Let the total numbers of the ball in a bag is x.
Blue balls = 2/3 of x and Pink ball = x - 2x/3 = x/3
Defective Blue balls = 5/9 of (2/3 of x) = 5/9 × 2/3 × x = 10/27 × x
Defective Pink balls = 7/8 of x/3 = 7x/24
Now, According to the Question, Non Defective Blue and Pink balls = 146
Total no. of balls=defective+Non defective
Non defective=Total -Defective
x - ( 10x/27 + 7x/24) = 146
x - ( 8(10x) + 9(7x) / 216 ) = 146
x - (80x + 63x)/216 = 146
x - 143x/216 = 146
x = 432

Total number of Student =100

Number of student passed first exam is 60, so probability is 60/100 = 0.6

Number of student passed second exam is 50/100, so probability is = 0.5

Number of student passed third exam is 30/100, so probability is = 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, probability of student passing at least in one subject is 0.8.

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.

 

The Probabilities of getting with replacement is =8/13*8/13*8/13=512/2197
THE probabilities getting without replacement=8/13*7/12*6/11=336/1716

First we have to select a bag and then we will draw a ball.

Probability of selection of both bags is equal =1/2

Now probability of blue ball taken from first bag = ( 1/2) x (3/11)

and probability of blue ball taken from second bag = (1/2) x (4/11)

So probability of blue ball = ( 1/2) x (3/11) + (1/2) x (4/11) = 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).

Explanation:
total probabilities for getting 5 = 4/36
total probabilities for getting 7 = 6/36
Total Probability = 10/36
We need only 5, hence probability of getting only 5 is (4/36)/(10/36)
=40%

Alternative way - (Difficult)

5 can be thrown in 4 ways (1,4), (2,3), (3,2), (4,1)

Hence the probability of throwing a 5 in a single throw with a pair of dice is 4/36 = 1/9

Similarly, 7 in 6 ways in a single throw with a pair of dice.

Hence the number of ways of throwing neither 5 nor 7 is 36 - ( 4 + 6 ) = 26

And the probability of throwing neither 5 nor 7 is ( 26/36 ) = ( 13/18 )

Hence the required probability

= ( 1/9 ) + ( 13/18 ) ( 1/9 ) + ( 13/18 )² ( 1/9 ) + ( 13/18 )³ ( 1/9 ) + .....

= ( 1/9 ) / [ 1 - ( 13/18 ) ] = 2/5 (using AP, GP formula)

EXPLANATION :

Since he has already thrown a 5, ( i.e a number different from 7), he may throw 5 at the next attempt. The probability for which is ( 1/9 ) or he may throw 5 at the second attempt when he fails to throw either 5 or 7 at the first attempt, the probability for which is ( 13/18 ) ( 1/9 ) or at the third attempt the probability for which is ( 13/18 )² ( 1/9 ) and so on ........ Adding all the probabilities we get the answer.

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

So let the letters be : A B C D E and the corresponding envelopes are : a b c d e.

Now A cannot go to a, so A can for example go to c.

Then C has 4 options to go to: a , b, d, and e. The 1st has 2 choices, the 2nd has 3, the 3rd has 3, and the 4th has 3 choices.

So there are 2 + 3 + 3 + 3 = 11 choices. Since A has 4 options to go to, the total is then : 11*4 = 44. And there are a total of 5! ways to do so.

So then probability is : 44/120.

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