TCS Ninja Questions on Probability Part 1

Probability of getting head = \frac{1}{2}

Probability of getting Tail = \frac{1}{2}

Probability of no head = \frac{1}{2}\times \frac{1}{2}\times\frac {1}{2} =\frac{1}{8}

Probabilty of getting at least one head = 1-\frac{1}{8} = \frac{7}{8}

n(A∪B)=60 +50 - 30= 80


it, is evident that 80 students passed at least 1 exam.

Thus, 20 failed both and the required probability is \frac{20}{100}=\frac{1}{5}.

The Probabilities of getting with replacement is =\frac{8}{13}\times \frac{8}{13}\times \frac{8}{13}=\frac{512}{2197}
The probabilities getting without replacement=\frac{8}{13}\times \frac{7}{12}\times \frac{6}{11}=\frac{336}{1716}

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

Probability of selection of both bags is equal = \frac{1}{2}

Now probability of blue ball taken from first bag = \frac{1}{2}\times \frac{3}{11}

and probability of blue ball taken from second bag = \frac{1}{2}\times \frac{4}{11}

So probability of blue ball =  \frac{1}{2}\times \frac{3}{11} + \frac{1}{2}\times \frac{4}{11}= \frac{7}{22}

P(Getting a red ball from the first bag): (\frac{1}{2})\times (\frac{5}{12})
P(Getting a red ball from the 2nd bag): (\frac{1}{2})\times (\frac{3}{15})
P(Drawing a red ball):
(\frac{1}{2})\times (\frac{5}{12})+(\frac{1}{2})\times (\frac{3}{15})= \frac{37}{120} (Ans).

ans: 40%
Explanation:
total probabilities for getting 5 = \frac{4}{36}
total probabilities for getting 7 = \frac{6}{36}
Total Probability = \frac{10}{36}
We need only 5, hence prob of getting only 5 is \frac{(\frac{4}{36})}{(\frac{10}{36})}
= 0.4

= 40%

Answer: c
Probability of thangam being selected= \frac{1}{3} and him not being selected = 1-\frac{1}{3}=\frac{2}{3}
Probability of pandiyamma being selected= \frac{1}{5}and him not being selected= 1-\frac{1}{5}= \frac{4}{5}
Probability that both are not selected = \frac{2}{3}\times \frac{4}{5}= \frac{8}{15}

 

Total number of ways = 5! =120

D_{n} = n!\sum_{k=0}^{n} = n!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}...+(-1)^{n}\frac{1}{n!})

= 5!\sum_{k=0}^{5} = 5!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}...+(-1)^{5}\frac{1}{5!})

= 44

Required Probability = \frac{44}{120}

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 x 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) x 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) x 10th place can be filled with 9 values (0 to 9 excluding 2, as we have already been counted 2 at 10th place) ]
=10 x 10+9 x 10+9 x 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 = \frac{290}{1100}

 



 

total events = 6 x 6 x 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 (\frac{3!}{2!} as repetition)
(3,3,4)-3 combinations
(2,2,6)-3 combinations
so total combinations are 27
so probability will be \frac{27}{216}