### CAT Online Coaching - Permutation and Combination, Fixing the Errors III

This post had given a series of
questions with incorrect solutions. Given below are the "debugged"
solutions to questions 7, 8 and 9.

7. What is the
probability of selecting 3 cards from a card pack such that all three are face
cards? what is the probability that none of the three is a face card?

**Given solution**

Number of cards in a
card pack = 52

Numbert of face cards
in a card pack = 12

Number of ways of
selecting 3 cards from a card pack = 52C3

Number of ways of
selecting 3 face cards from a card pack = 12C3

Probability of
selecting three cards such that all three are face cards = 12C3/52C3

Probability of
selecting three cards such that none of the three are face cards = 1 -
12C3/52C3

**Bug in the solution:**

Other possibilities
exist. As in, if we select three cards from a card pack, all three could be
face cards, all three could be non-face cards, one could be a face card with 2
non-face cards or we could have two face cards and one non-face card. This is
why we cannot use the 1 minus idea.

As a rule we can say
P(A) = 1 – P(B) – P(C) if A, B and C are mutually exclusive and collectively
exhaustive events. As in among them they should account for all possible events.
And there should be no overlap. Creating a group of MECE equiprobable events is
the most fundamentally brilliant idea in all of probability. Go on, look it up.

**Correct solution:**

This is simple. Probability
of selecting three cards such that none of the three are face cards = 40C3/52C3

8. A die is rolled
thrice. In how many outcomes will two throws be same and the third one
different?

**Given solution**

Let the three
outcomes be ABC.

'A' can take all
values from 1 to 6

'B' can also take all
values from 1 to 6

'C' can take all
values except A - so it has 5 possibilities

Total number of
outcomes = 6 * 6 * 5 = 180.

**Bug in the solution:**

The given solution is
actually absurd. If two throws are to be same, then if A can take values from 1
to 6 and if B were equal to A, then b can take only one value. There can be no
6 * 6 * 5

**Correct solution:**

'A' can take all
values from 1 to 6

'B' should be equal
to A --- One possibility

'C' can take all
values except A - so it has 5 possibilities

6 * 1 * 5 = 30
outcomes.

There are 30 possible
outcomes when A = B but not equal to C. Likewise, we could have A = C but not
equal to B and B = C but not equal to A. So, there are totally 90 different
possibilities.

9. How many 7 letter
words can we have in English that have two distinct vowels and 5 distinct
consonants.

**Given solution**

Now, we know there
are 5 vowels and 21 onsonants. So, we need to select 2 from these 5 and 5 from
the remaining 21. Since order is important, we need to select keeping order in
mind.

So, we
have 5P2 * 21P5.

**Bug in the solution:**

In this solution we
do not account for intermingling of vowels and consonants. As in, we do not
count words such as BACEDFG. We account for order within vowels and order
within consonants, but we do not account for order across both categories.

**Correct solution:**

Select without
accounting for order, then arrange everything put together

So, we
have 5C2 * 21C5 * 7!.

Labels: CAT Coaching classes, CAT Coaching Online, CAT preparation, CAT Preparation online

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