This is the aptitude questions and answers section on 'Probability' with solutions and detailed explanation. Questions on the basic definitions and theorems of probability are asked.
Problems on Probability : Question 1 :
If two dice are thrown simultaneously, then find the probability that the sum of numbers appeared on the dice is 6 or 7?
The sum of numbers appeared is 6 or 7. Therefore, the required sums are 6 or 7, i.e., the required events are (1,5), (5,1), (2,4), (4,2), (3,3), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)
i.e., for 6, n(E) = 5 and for 7, n(E) = 6
Therefore, the required probability = n(E)/n(S) = 11/36.
Problems on Probability : Question 2 :
The probability that Alex will solve a problem is 1/5. What is the probability that he solves at least one problem out of 4 problems?
The probability that Alex will not solve a problem = 4/5. The probability that Alex will not solve 10 problems = (4/5)4
= 256/625. Hence, the probability that Alex will solve at least one problem = 1 - 256/625 = 369/625.
Problems on Probability : Question 3 :
In a race where 12 cars are running, the chance that car X will win is 1/6, that Y will win is 1/10 and that Z will win is 1/8. Assuming that a dead heat is impossible, find the chance that one of them will win.
Required probability = P(X) + P(Y) + P(Z) (all the events are mutually exclusive).
= 1/6 + 1/10 + 1/8 = 47/120
Problems on Probability : Question 4 :
Out of 17 applicants, there are 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.
The events of selection of two person is redefined as (i) first is a girl AND second is a boy OR (ii) first is boy AND second is a girl OR (iii) first is a girl and second is a girl.
So the required probability:
= (8/17 x 9/16) + (9/17 x 8/16) + (9/17 x 8/16)
= 9/34 + 9/34 + 9/34
Problems on Probability : Question 5 :
What is the probability that in the rearrangements of the word AMAZED, the letter ’E’ is positioned in between the 2 ’A’s (Not necessarily flanked)?
In any rearrangement of the word, consider only the positions of the letters A, A and E.
These can be as A A E, A E A or E A A.
Therefore, effectively one-third of all words will have ’E’ in between the two ’A’s.