Probability Questions Problems

probability questions in TCS solved Q's: 

Question 1

Four friends namely Rahul, Ravi, Rajesh and Rohan contested for a dairy milk chocolate. To decide which friend will get the chocolate they decided to throw two dice. Every friend was asked to choose a number and if the sum of the numbers on two dice equals that number, the concerned person will get the chocolate. Rahul's choice was7, Ravi's choice was 9, Rajesh's choice was 10 and Rohan's choice was 11. Who has the maximum probability of winning the amount.

a) Rahul b) Ravi c) Rajesh d) Rohan

Answer : a) Rahul

Solution :

Number 7 will appear more often –(1,6), (2,5), (3,4), (4,3), (6,1), (5,2) --- 6 cases
Number 9 -- (3,6),(6,3), (4,5) (5,4) ---4 cases
For number 10 -- (4,6) ,(6,4) (5,5) ---3 cases
For number 11 -- (5,6),(6,5)...2 cases.
Since number 7 has the maximum chance of appearing, it will have the maximum probability as well. Hence, Rahul will most probably be the winner.

Question 2

Yegaraj, Zimbo, Kodanda, Ramiah and Kashi were given undescribed measure of gold coins, silver bars etc. Differences arose among them as to how these are to be distributed. Different strategies/options/methodologies were discussed. Finally it was decided to throw two dice. Every friend was asked to choose a number. Assume Yegaraj chooses 9 and Zimbo chooses 10. By how much percentage, probability of Yegaraj winning is more than that of Zimbo. 

a) 30% b)66.67% c)33.33% d) 25%

Answer : c) 33.33%

Solution :

Possible cases for the sum of the numbers to be 9 -- (3,6),(6,3), (4,5) (5,4) ---4 cases
Possible cases for the sum to be 10 -- (4,6) ,(6,4) (5,5) ---3 cases
Total number of combinations when two dice are thrown = 36 (... 6 numbers on first die X 6 numbers on the second die)
P1 = probability of Yegaraj winning = Possible cases for the sum to be 9 / Total combinations = 4/36
P2 = probability of Zimbo winning = Possible cases for the sum to be 10 / Total combinations = 3/36
Percentage by which P1 is more than P2 = (P1 - P2) / P2 x 100% = (4/36 - 3/36) / 3/36 x 100% = 1/3 x 100% = 33.33%

Question 3

A professor of Statistics posed the following question to his students:
“Three dices (six sided) are thrown one after the other. What is the probability of getting a different number on each dice?”

a) 1/12 b) 1/3 c) 5/9 d) 4/9

Answer : c) 5/9

Solution :

Dice rolled first can show any of the six numbers – so there are six possibilities. Second dice shall not have the number rolled by first dice. So there are 5 possibilities. Similarly for third dice has four possibilities—i.e. it should not show the number shown by first dice and/or second dice.
Based on the above argument, Number of possibilities for three dice to show different numbers = 6 x 5 x 4 = 120.
Also we know that the total possibilities when three dice are thrown = 6 x 6 x 6 = 216
Probability of getting a different number on each dice = Number of Possibilities for three dice to show different numbers / Total Possibilities = 120/216 = 5/9


Question 4

Two countries - Germany and France are participating in a hockey game. Mr. Randor Guy, the famous astrologer in Germany is able to predict the winner of each match with great success. It is rumoured that in a match between 2 teams X and Y,Rando Guy picks X with the same probability as X's chances of winning. Let's assume such rumours to be true and that in a match between Germany and France, Germany the stronger team has a probability of 7/10 of winning the game. What is the probability that Randor Guy will correctly pick the winner of the Germany - France game?

a) 0.58 b) 0.68 c) 0.82 d) 0.42

Answer : a) 0.58

Solution :
Randor Guy predicts the winner in games with the same probability of winning. --> statement I
Germany has got a chance of winning the game with 7/10 probability. France has got a chance of winning the game with 3/10 probability.
Probability of Randor Guy picking the Germany correctly is 7/10 while picking France is 3/10 (as per statement I).
Probability of Randor Guy picking the winning team correctly = (Probability of Germany winning the game) AND (Probability of Randor Guy picking Germany correctly) OR (Probability of France winning the game) AND (Probability of Randor Guy picking France correctly).
Note : In probability related formulas generally AND translates to x and OR translates to +
Therefore, Probability of Randor Guy picking the winning team correctly = (Probability of Germany winning the game) * (Probability of Randor Guy picking Germany correctly) + (Probability of France winning the game) * (Probability of Randor Guy picking France correctly).
Probability of predicting winning team = (7/10 * 7/10) + (3/10 * 3/10)
=0.58

Question 5

A box has 45 chocolates of different colours 20 red, 15 white and 10 black chocolates. If a chocolate is chosen at random, what is the probability of that being white chocolate?

a) 1/2 b) 1/3 c) 2/3 d) 3/4

Answer : b) 1/3

Solution :
Note : This is a simplest of the simple questions that you can expect on probability. This is a usual case in placement tests. You should be able to identify such simple questions amidst tough questions. At times, you could get mislead by thinking this is a tricky one.
Any of 45 chocolates could be chosen. So, totally there are 45 possibilities. Since there are only 15 white chocolates, only 15 possibilities can lead to a white chocolate. Hence P(white chocolate) = 15/45 or 1/3.

Question 6

In an off campus placement programme, a software company recruiter interviewed 75 prospective candidates -- 10 from Civil engineering department, 5 from Bio-chemical department and rest from computer science department. If the software company finally issued offer letter to 17 candidates, what is the probability that all the selected candidates belonged to only computer science department?

a) 60C17/ 75C17 b) 65C17/75C17 c) 70C17/75C17 d) None of these.

Answer : a) 60C17/75C17

Solution :
17 out of 75 candidates selected in 75C17 ways. 17 out of 60 from computer science department can be selected in 60C17 ways.
So the required probability = Ways of selecting 17 computer science candidates / Ways of selecting 17 candidates from all departments
= 60C17 / 75C17