- 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.

2/354/356/355/353/35Option 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 - 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
1/401/301/2001/101/100Option 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 - Six boys and five girls stands in queue for buy a pizza in a shop. The probability that they stand in alternate positions is:
1/4521/4621/2621/3621/162Option 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 - 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?
119125110115120Option B

(4C1 * 5C4 ) + (4C2 * 5C3) +(4C3 *5C2) +(4C4 *5C1) =60+60+40+5

=125 - 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?
335300320330310Option D

1 is permanent. We have to choose other from 11

=11C4 = 330 - 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 /1209900 /1309100 /1309800 /1309700 /1309Option B

basket should contain one apple and 2 oranges = 10C2 *25C1 *32C1

Total probability =35C4

So the required probability =900 /1309 - How many groups of 6 students can be formed from 8 boys and 7 girls ?
40055005600520053005Option 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 - How many different ways the letters can be formed “JOBUPDATES” so that all the vowels never come together?
36445203600520363352036115203622520Option D

All the come together = 6! 4!

Total choices = 10!

All vowels never come together =10! -6! 4! =3611520 - 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 ?
259278286216206Option C

4C3 +6C3 +3C3 =25

13C3 =286 - 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?
4/226/2211/229/2210/22Option D

Probability = 3c1 *9c1 / 12c2 = 3*9 /66 = 9/22

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