My friend Parveen teaches at a school. One day she conducted a test for three of her students and when they handed back the test papers, they had forgotten to write their names.

Parveen returned the papers to students at random.

What is probability that none of three students will get the right paper?

Please follow and like us:

3 students so 3 papers so possibility of

1. correct and

2. not correct are the possibilities for given scenarios in question.

now: 2 type of possibilities per student, hence 2 ^ 3 = 8 possibilities for total 3 students

hence our N = 8

now favorable case is – 0 for none of total students get correct answer sheet back

I would do this based on a shortcut key:

11 ^ 3 = 1 3 3 1

———————————–

positions 0 1 2 3 — favorable events

for given scenario, favorable even makes as 0 so its related digit is -> 1

hence….

1/N = 1/ 8

answer will be 1/8 for given question above.

2/3 that first student gets wrong paper

1/2 that second student gets wrong paper ( assuming first gets wrong paper )

1 that third student gets wrong paper ( assuming first and second get wrong paper )

So probability = 2/3 * 1/2 * 1 = 1/3

1-1/3=2/3.

1/6

what are the 2 favourable cases?

abc,acb,bac,*bca,*cab,cba

asterix marked are condition

is the ans. 2/3.?????????

2/3

total cases – 6

favourable cases – 2

probability – 2/6 = 1/3