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Probability Level 5

In maths paper there is a question on "match the columns" in which column A contains $6$ entries and each entry of column A corresponds to exactly one of the $6$ entries given in column B written randomly $2$ marks are awarded for each correct matching and $1$ mark is deducted for each incorrect matching. I, having no subjective knowledge decides to match all $6$ entries randomly . Find the number of ways in which I can answer to get at least $25$ % marks in this question.

Source: AITS

The answer is 56.

1 solution

Parth Lohomi
Nov 18, 2014

Maximum marks in the particular question =12

25% marks means =12× $\dfrac{25}{100}$ =3

$case :1$

When I get 3 marks i. e 3 correct and 3 wrong answers

6 $C_{3}$ ×3![1- $\dfrac{1}{1!}$ - $\dfrac{1}{2!}$ - $\dfrac{1}{3!}$ ]= $\dfrac{6×5×4}{1×2×3}$ [ $\dfrac{1}{2}$ - $\dfrac{1}{6}$ ]×6= $40$

$case :2$

When $4$ are correct and $2$ wrong

6 $C_{4}$ ×2![1- $\dfrac{1}{1!}$ + $\dfrac{1}{2!}$ ]= $15$ ways

$case :3$

When $5$ correct and $1$ wrong

6 $C_{5}$ ×1!(1-1)=0….......(not possible)

$case :4$

all $6$ are correct

6 $C_{6}$ = $1$

Hence total ways= $40+15+1$ = $\boxed{56}$

Please upvote if you are satisfied

Parth Lohomi - 6 years, 6 months ago

i have a query, your solution is correct but I thought of this problem as:

select 3 choices which must be correct: 6C3 now it doesn't matter what goes with other 3 options, hence it should be any matching of 3 LHS options to RHS ones. that is 3!

so it should be 6C3 * 3!

Where am I wrong?

SHIV GUPTA - 6 years, 6 months ago

This would result in over-counting. For example, if it ended up that all 6 were matched correctly, then your counting method would end up counting this outcome 6C3 = 20 times.

Brian Charlesworth - 6 years, 3 months ago

@Brian Charlesworth – This is similar to finding number of solutions of 2x-y>=3 where 0<=(x,y)<=6. And when we find it, the no. of solutions comes out to be 26. Please correct me

Vighnesh Raut - 6 years, 1 month ago

Nice question, Parth. In the case of getting $3$ correct, and thus $3$ wrong, once the $\binom{6}{3}$ factor is taken care of we have a derangement of the final $3$ entries, which can be done in $2$ ways.

This is a good question to create variations of. For example, if there were $10$ entries in each column and we wanted a score of at least $25$ %, then the number of ways would be

$\binom{10}{5}*44 + \binom{10}{6}*9 + \binom{10}{7}*2 + \binom{10}{8} + \binom{10}{10} = 13264$ .

Brian Charlesworth - 6 years, 6 months ago

Nice question with nice solution. Great !

Sandeep Bhardwaj - 6 years, 6 months ago

Thanks Sir!!

Parth Lohomi - 6 years, 6 months ago

We have 720 ways in total.

First i was doing this ques. the reverse way i.e subtracting the no. of ways out of 720 when getting atleast 25% is not possible.

but i got a way bigger answer than the real one.

WHY !

Mukul Sharma - 5 years, 10 months ago

Please let me pass!

The answer is 56.

1 solution

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