Question:

Practice Questions on Probability !
The maths question contains atleast one question on probability. I have been receiving queries from the students who are facing problem in probability. To help all learn-hubbers, we will posting few questions on probability every week here.

The probability of event A is the number of ways event A can occur divided by the total number of possible outcomes.

Probability of event A =

(No of ways event A can occur ) / ( Total number of possible outcomes )

Helpful Hints about Probability

1. If an event is certain to occur, the probability is 1.

for e.g : An event with a probability of 1 can be considered a certainty: for example, the probability of a coin toss resulting in either "heads" or "tails" is 1, because there are no other options, assuming the coin lands flat.

2. If an event is certain not to occur, the probability is 0.

An event with a probability of 0 can be considered an impossibility: for example, the probability that the coin will land (flat) without either side facing up is 0, because either "heads" or "tails" must be facing up.

3. If you know the probability of all other events

occurring, you can find the probability of the

remaining event by adding the known probabilities

together and subtracting from 1.

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Sample Question : A single 6-sided die is rolled. What is the probability of each outcome?

The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6.

Probabilities:

P(1) = (No. of ways to roll a 1) / Total number of possible outcomes = 1 / 6

P(2) = (No. of ways to roll a 2) / Total number of possible outcomes = 1 / 6

P(3) = (No. of ways to roll a 3) / Total number of possible outcomes = 1 / 6

P(4) = (No. of ways to roll a 4) / Total number of possible outcomes = 1 / 6

P(5) = (No. of ways to roll a 5) / Total number of possible outcomes = 1 / 6

P(6) = (No. of ways to roll a 6) / Total number of possible outcomes = 1 / 6

Getting 2,4,6 are possible even outcomes.

P(getting an even number) = 3/6 = 1/2

Practice Question 1:

Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?

(A) 1 / 18 (B) 64 / 4032 © 63 / 64 (D) 1 / 9 (E) 2 / 19Which one is the correct answer?

Keep checking this page for more practice questions and solutions !

Hey, have you tried to solve this question by yourself. If yes, read on further, else, I would suggest you to first try your hands on it and then move further…This is the best way to improve your skills.

The correct answer is A.

Solution:

Probability of having a side in common = (Favorable Cases) / (Total number of cases)

First, let us find out total number of ways of choosing two
squares on chessboard. There are total 64 squares on a chessboard,
so the number of ways of choosing the first square is 64. The
number of ways of choosing the second square is 63.

There are a total of 64 * 63 = 4032 ways of choosing two
squares.

If the first square happens to be any of the four corner ones, the second square can be chosen in 2 ways. If the first square happens to be any of the 24 squares on the side of the chess board, the second square can be chosen in 3 ways. If the first square happens to be any of the 36 remaining squares, the second square can be chosen in 4 ways.

Hence the desired number of combinations = (4 * 2) + (24 * 3) + (36 * 4)

= 224.Therefore, the required probability

= 224/4032

= 1/18 .

consider the squares along the sides…u say that ways of getting an adjacent sq is 3.but when u concider the inner squares (for the sq's just adjacent to the ones along the sides) arent u once again considering the same combination by saying that the chance of getting an adjacent sq is 4

Practice Problem 2:

A shooter can hit a target once in 4 shots. If shooter fires 4 shots in succession, what is the probability that the shooter will hit his target?

A. 3/4

B. 1

C. 1/256

D. 81/256

E. 175/256

Post your replies.

y dunt we take 1/4*1/4*1/4*1/4 ples xplain

Yes, that is the correct answer.

Here is the solution :

The shooter will hit the target even if he hits it once or twice or thrice or all four times in the four shots that

he takes.

So, the only case where the shooter will not hit the target is when he fails to hit the target even in one of the

four shots that he takes.

The probability that he will not hit the target in one shot = 1
- 1/4 = 3/4

Therefore, the probability that he will not hit the target in all
the four shots = 3/4 * 3/4 * 3/4 * 3/4 = 81/256

Hence, the probability that he will hit the target at least in
one of the four shots = 1 - 81/256

= 175/256 .

Here is the next question :

When two dice are thrown simultaneously, what is the probability that the sum of the two numbers is less than 11?

A. 5/6

B. 11/12

C. 1/6

D. 1/12

E. 2/3

Post your replies.

That is correct.

Here is the detailed solution :

Instead of finding the probability of this event directly, we
will find the probability of the non-occurrence of this event and
subtract it from 1 to get the required probability.

Total possible combinations : 6*6 = 36

Combination whose sum is 12 : (6,6)

Combinations whose sum is 11 : (5,6), (6,5).

Therefore, there are totally 3 occurrences out of 36 occurrences that satisfy the given condition.

Probability whose sum of two numbers is greater than or equal to 11 = 3 / 36 = 1 / 12.

Hence probability whose sum of two numbers is lesser than 11 = 1 - 1 / 12 = 11 / 12.

yes..it is b..please keep posting more questions

Click here for the solution of above question.

Here is the next question:

The probability that John speaks truth is 3/4 and the probability for Cheryl speaking truth is 4/5. What is the probability that they contradict each other when asked to speak on certain fact ?

A) 3/20

B) 2/5

C) 7/20

D) 1/5

E) 3/5

That is correct.

Solution :

Let E1 be the event denoting that M speaks truth and E2 be the
event denoting that N speaks truth.

Probability that both contradict each other :

Probability that John speaks truth = P(J)= 3/4

Probability that John lie : 1/4

Probability that Cheryl speaks truth= P©= 4/5

Probability that Cheryl lie : 1/5

Probability that both contradict each other :

= (3/4) * (1/4) + (1/5) * (3/4)

= 7/20

Click here for solution of above question.

Practice Problem 5

A bag has 10 black balls and 10 White balls. Person A picks two balls at random, than Person B picks two of the remaining balls at random. What is the probability of them getting the same color combination.

A. 118 / 323

B. 121 / 323

C. 81 / 323

D. 205 / 323

E. 181 / 323

Answer : A

Solution :

If Person A makes the first two picks and Person B makes the last two picks, there are six different combinations that work.

BBBB, WWWW, BWBW, BWWB, WBBW, WBWB

Chances of each of these scenarios occurring are:

BBBB = 10 / 20 * 9 / 19 * 8 / 18 * 7 / 17 = 5040 / 116280

WWWW = 10 / 20 * 9 / 19 * 8 / 18 * 7 / 17 = 5040 / 116280

BWBW = 10 / 20 * 10 / 19 * 9 / 18 * 9 / 17 = 8100 / 116280

BWWB = 10 / 20 * 10 / 19 * 9 / 18 * 9 / 17 = 8100 / 116280

WBBW = 10 / 20 * 10 / 19 * 9 / 18 * 9 / 17 = 8100 / 116280

WBWB = 10 / 20 * 10 / 19 * 9 / 18 * 9 / 17 = 8100 / 116280

Adding these all together, we get the desired probability = 42480 / 116280 = 4248/11628 = 1062/2907 = 354/969 = 118/323 = 36.53%

Click here for solution of above problem.

Practice Problem 6

An unbiased die is thrown 6 times. What is the probability that the numbers will have a product which is an even number.

Answer 2

Probability of scores with even product = 1 - (probability of all
odd numbers)

= 1 - [ ( 1 / 2 ) ^ 6 ]

= 63 / 64

Total number of ways=6^6

To get the answer first we have to calculate no of ways to get the
product as ODD. if 2 or 4 or 6 are there
in any of the throws,we wont get an ODD
number.So exclude those three(2,4,6) remaining three(1,3,5) only
should be there. so in every throw any of these three(1,3,5) will
come. No.of ways to get odd product is 3.3.3.3.3.3=729

so probability of getting odd o/p is 729/6^6

probability of getting even product=1-(1/64)=63/64

explanantion :

the probability of getting a odd number when the dice is thrown
once is (1/2).

similarly when the dice is thrown 6 times, the probabilty of
getting odd every time is (1/2) ^6.

so reqd probability = 1 - [ ( 1 / 2 ) ^ 6 ]

= 63 / 64 .

Practice Problem 7

Ray plays a game involving throw of 2 unbiased dice where the person throwing a total of 7 or 11 is the winner. What is the probability of Ray winning the game.

A. 1/9

B. 2/9

C. 4/9

D. 5/9

E. None of these

Hello Can someone here can post Five Problems Each of the
Following:

2. Probability if mutually Exclusive Events

3. Probablity of Events that are not Mutually Exclusibve

4. Probability with Independent Events

5. Dependent Events

Thanks Alot :)))

total how many questions is there in test

by **kotiprince**

Preparing from the scratch?

by **trsram**

No Internship/ research experience. Chances of getting Grade A Universities?

by **priyanka1591**

need help in university selection

by **suvaa**

HI.. My GRE score is 303 . Quant - 160/170 verbal 143/170

by **maverick_87**

Gre score 311, quant : 164, verbal : 147 ? any chances

by **mnarang**

total how many questions is there in test

More Questions
by kotiprince
Sat, 06 Oct 2012 07:07:15 -0000

Preparing from the scratch?
by trsram
Thu, 04 Oct 2012 16:55:43 -0000

No Internship/ research experience. Chances of getting Grade A Universities?
by priyanka1591
Tue, 02 Oct 2012 20:42:57 -0000

need help in university selection
by suvaa
Tue, 02 Oct 2012 05:43:02 -0000

HI.. My GRE score is 303 . Quant - 160/170 verbal 143/170
by maverick_87
Sun, 30 Sep 2012 11:13:49 -0000

Gre score 311, quant : 164, verbal : 147 ? any chances
by mnarang
Fri, 28 Sep 2012 13:27:11 -0000

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