Thursday, December 19, 2019

Quantitative Aptitude: Probability Questions – Set 12

  1. Two friends A and B appeared for an selection . Let E1 be the event that A is selected and E2 is the event that B is selected. The probability of E1 is 2/5 and that of E2 is 3/7. Find the probability that both of them are selected.
    2/35
    4/35
    6/35
    5/35
    3/35
    Option C
    Given, E1 be the event that A is selected and
    E2 is the event that B is selected.
    P(E1)= 2/5
    P(E2)=3/7
    Let E3 be the event that both are selected.
    P(E1)=P(E1)×P(E2) as E1 and E2 are independent events:
    P(E3) = 2/5*3/7
    P(E3) =6/35
    The probability that both of them are selected is 6/35

     


  2. A policeman forgot the last digit of an 11 digit land line phone number. If he randomly dials the final 2 digits after correctly dialing the first nine, then what is the chance of dialing the correct number

    1/40
    1/30
    1/200
    1/10
    1/100
    Option E
    It is given that last two digits are randomly dialed.
    Then each of the digits can be selected out of 10 digits in 10 ways.
    Hence required probability
    =(1/10)2
    =1/100


     


  3. Six boys and five girls stands in queue for buy a pizza in a shop. The probability that they stand in alternate positions is:

    1/452
    1/462
    1/262
    1/362
    1/162
    Option B
    Total number of possible arrangements for Six boys and five girls stand in queue =11!
    When they occupy alternate position the arrangement would be like:
    bgbgbgbgbgb
    Thus, total number of possible arrangements for boys,
    = 6*5*4*3*2
    Total number of possible arrangements for girls,
    =5*4*3*2
    Required probability
    =6*5*4*3*2*5*4*3*2/11*10*9*8*7*6*5*4*3*2
    = 1/462


     


  4. From 4 roses and 5 jasmines , the garland has to be formed with 5 flowers and it contain at least one rose. In how many ways the garland can be formed?

    119
    125
    110
    115
    120
    Option B
    (4C1 * 5C4 ) + (4C2 * 5C3) +(4C3 *5C2) +(4C4 *5C1) =60+60+40+5
    =125

     


  5. There are 12 members in a committee. If a group of 5 members has to be selected from the committee in such way that one member is always included, in how many different ways the selection can be done?

    335
    300
    320
    330
    310
    Option D
    1 is permanent. We have to choose other from 11
    =11C4 = 330

     


  6. The ratio of apples and oranges is 5:2 and total count is 35 .A basket of 4 fruits should be taken. What is the probability that the selected group contains 1 apple and 2 oranges?

    900 /1209
    900 /1309
    100 /1309
    800 /1309
    700 /1309
    Option B
    basket should contain one apple and 2 oranges = 10C2 *25C1 *32C1
    Total probability =35C4
    So the required probability =900 /1309


     


  7. How many groups of 6 students can be formed from 8 boys and 7 girls ?
    4005
    5005
    6005
    2005
    3005
    Option B
    Total no . of students = 8 + 7 = 15
    No. of groups = 15C6 = 15! / {6! (15 - 6)!} = 15! / (6! 9!)
    = (15 x 14 x 13 x 12 x 11 x 10) / (6 x 5 x 4 x 3 x 2 x 1)
    = 5005


     


  8. How many different ways the letters can be formed “JOBUPDATES” so that all the vowels never come together?
    3644520
    3600520
    3633520
    3611520
    3622520
    Option D
    All the come together = 6! 4!
    Total choices = 10!
    All vowels never come together =10! -6! 4! =3611520

     


  9. In a box there 4 marbles in blue , 6 marbles in yellow, 3 marbles in red color and we have to take 3 marbles and what is the probability taking all marbles in same colour ?

    259
    278
    286
    216
    206
    Option C
    4C3 +6C3 +3C3 =25
    13C3 =286


     


  10. A box contains dozen of chocolates out which 3 are milk rest are cocoa. Two chocolates are taken out. What is the probability of one chocolate is milk and another one is cocoa?

    4/22
    6/22
    11/22
    9/22
    10/22
    Option D
    Probability = 3c1 *9c1 / 12c2 = 3*9 /66 = 9/22

     




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