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Practice Questions on Probability !

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Asked by deepa mittal

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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    deepa mittal
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    Deepa MittalFri, 22 Jan 2010 08:01:12 -0000

    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

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    deepa mittal
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    Deepa MittalFri, 22 Jan 2010 08:08:17 -0000

    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 / 19

    Which one is the correct answer?

    Keep checking this page for more practice questions and solutions !

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    deepa mittal
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    Deepa MittalMon, 25 Jan 2010 04:15:03 -0000

    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 .

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    shravansriram
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    Shravan SriramThu, 02 Jun 2011 05:00:16 -0000

    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

    Pavan2310
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    Pavan2310Wed, 03 Oct 2012 08:56:23 -0000

    I think the asnwer is 121/4032.. Let me know if someone comes with same answer, then I'll explain the logic I used to solve.

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    deepa mittal
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    Deepa MittalMon, 25 Jan 2010 04:18:16 -0000

    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.

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    some21
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    Soma DuttaSat, 30 Jan 2010 04:11:41 -0000

    answer is 175/256.

    Chintu_sathwik
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    Sathwik ReddyMon, 23 Jul 2012 19:26:57 -0000

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

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    deepa mittal
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    Deepa MittalSat, 30 Jan 2010 04:13:36 -0000

    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 .

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    deepa mittal
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    Deepa MittalSat, 30 Jan 2010 04:18:22 -0000

    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.

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    santhoshdayalan
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    Santhosh RajadayalanTue, 02 Feb 2010 11:09:49 -0000
    deepa mittal
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    Deepa MittalSat, 06 Feb 2010 11:27:25 -0000

    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.

    angelpreeti
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    Manpreet KaurSun, 07 Mar 2010 12:16:10 -0000

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

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    deepa mittal
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    Deepa MittalSat, 06 Feb 2010 11:32:24 -0000

    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

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    ofayyaz
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    ofayyazSun, 07 Feb 2010 13:54:30 -0000

    (Prob John (true) x prob cher (false)) + (prob cher (true) x prob john (false)) = 7/20 ©

    deepa mittal
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    Deepa MittalMon, 22 Feb 2010 11:40:18 -0000

    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

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    deepa mittal
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    Deepa MittalMon, 22 Feb 2010 11:47:33 -0000

    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

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    deepa mittal
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    Deepa MittalThu, 25 Feb 2010 05:15:36 -0000

    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%

    wajalie
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    wajalieTue, 16 Oct 2012 14:21:43 -0000

    why wasn't the combination BBWW and WWBB considered.

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    deepa mittal
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    Deepa MittalThu, 25 Feb 2010 05:17:17 -0000

    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.

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    deepa mittal
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    Deepa MittalSun, 28 Feb 2010 18:03:58 -0000

    Answer 2
    Probability of scores with even product = 1 - (probability of all odd numbers)

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

    = 63 / 64

    angelpreeti
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    Manpreet KaurSun, 07 Mar 2010 12:26:59 -0000

    please explain this.

    ravichegondi
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    ravichegondiMon, 16 Jul 2012 10:41:00 -0000

    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

    vinay_shasthri
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    Vinay ShasthriMon, 24 Sep 2012 06:41:58 -0000

    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 .

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    deepa mittal
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    Deepa MittalSun, 28 Feb 2010 18:05:39 -0000

    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

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    tinamary_A
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    tina maryWed, 21 Apr 2010 01:35:26 -0000
    Gyanshu
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    GyanshuThu, 06 Jan 2011 17:31:52 -0000
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    lashort
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    lashortMon, 05 Nov 2012 09:59:01 -0000

    you have a bag containing 5 beads of which 2 are blue, 2 are red and 1 is green.You pick 3 beads at random without replacement.find the probability that you will get;
    2 blues and 1 red
    atleast a green and a blue
    atleast 1 red
    no green

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    HyunRiNeedsToLearn
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    HyunRiNeedsToLearnMon, 31 Dec 2012 05:25:29 -0000

    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 :)))

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