Below are three probability questions for practice.
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
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.
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%
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%
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
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