Lesson Typical probability problems from the archive

Typical probability problems from the archive

Problem 1

The probability that he will attend Math class    = 1 - 0.24 = 0.76.


The probability that he will attend Biology class = 1 - 0.11 = 0.89.


The probability that he will attend both classes  = 0.69    ( ! given ! )


The probability that he will attend at least one of the two classes =

    0.76 + 0.89 - 0.69 = 0.96.    (*)


The probability that he will skip both classes is the complement to (*)

    1 - 0.96 = 0.04.


Answer.  The probability under the question is  0.04.

Problem 2

For each single system of five, the probability that it will not detect a theft is  the complement  to 0.8,  i.e.  1-0.8 = 0.2.


The probability that NO ONE of five systems will detect a theft is   = 0.00032.


The probability that when a theft occurs, at least one of the 5 systems will detect it is THE COMPLEMENT to it,  i.e.  1 - 0.00032 = 0.99968.

Problem 3

1.  44 + 33 + 44 = 121.



2.  The probability to get two red balls     = .

    
    The probability to get two white balls   = .


    The probability to get two green balls   = .



3.  The probability to get two balls of the same color is  


      =  =  =  =  =  = 0.3333 = 33.33%.

Problem 4

Wanted outcomes are 3, 6, 9, 10, 11, 12.  


3 = 1+2, or 2+1,                ====>  so there are 2 ways to get the sum of 3.


6 = 1+5, 2+4, 3+3, 4+2, or 5+1  ====>  so there are 5 ways to get the sum of 6.


9 = 6+3, 5+4, 4+5, 3+6          ====>  so there are 4 ways to get the sum of 9.


10 = 6+4, 5+5, 4+6              ====>  so there are 3 ways to get the sum of 10.


11 = 6+5 and 5+6                ====>  so there are 2 ways to get the sum of 11.


12 = 1+1                        ====>  so there is  1 way  to get the sum of 12.


Thus you have  2 + 5 + 4 + 3 + 2 + 1 = 17 wanted cases of 36 possible outcomes.


The probability under the question is  .

Problem 5

The Probability = the ratio (of the number of the flights A left on-time) to (the number of all flights left on-time) = 


     = 0.52 = 52%.

Problem 6

The probability that ALL THREE missiles will hit target is the product


    0.5*0.7*0.4 = 0.14.


The probability that at least one missile will not hit the target is the COMPLEMENT to it, i.e.  1 - 0.14 = 0.86.

Problem 7

Answer.   =  = .

How many triples of man can be formed of 9 man ? -  =  = 84.


How many triples of women can be formed of 7 women ? -  =  = 35.


Hence, the entire space of events under the question is  84 + 35.


Of them, only 84 are favorable.


It gives you the answer.

Problem 8

You can select 7 manufactures from 11 manufactures by   ways =  = 330 ways.


It is your sample space, or space of events.


How many of them contain exactly 2 good of the best 3?


First, you can select two good manufactures from the best three by 3 ways.


Second, you can complement these 2 manufactures to 7 by adding 5 manufactures from 11-3 = 8 remaining manufactures,

and you can do it by   ways =  = 56 ways.


Thus 3*56 is the number of favorable choices.


Then the probability under the question is   = 0.5091 = 50.91%.    ANSWER

Problem 9

    In all, there are   =  = 6 groups of 2 students, that can be formed of 4 students.

    Of these 6 groups, exactly 3 groups contain "3".  They are  (3,1), (3,2), and (3,4).

    Thus the probability under the question a) is   = .

    There is ONLY ONE group of two students, which contain NEITHER "3"  NOR  "4".

    This group is (1,2).

    The rest 5 of 6 groups necessary contain  EITHER "3"  OR  "4".

    Thus the probability under the question b) is  .

    As we saw in the n. a), there are 3 groups of two students that contain "3".

    Hence, of the total possible 6 groups three contain "3", while 3 other do not.

    Thus the probability under the question c) is   = .