This post had given a series of questions with incorrect solutions. Given below are
the "debugged" solutions to questions 4, 5 and 6. The questions and the incorrect solutions here
4. In how
many ways can we select 2 cards from a card pack such that both belong to the
card can be selected in 52 ways, as it can be any card in the pack. Now, there
are 12 different ways of selecting the second card. Depending on the suit of
the first card, the choice set we have for the second card gets limited. If the
first card were a club, then the second card must be from among the 12
remaining clubs and so on.
number of outcomes = 52 * 12.
double-counting when we do this. If we select 2H as the first card and 5D as
the second card; this is same as selecting 5D as the first card and 2H as the
second card. However, in our current approach, we will end up counting both of
correct approach would be to select one of the four suits first and then select
2 cards from 13 in that suit. Or, the number of ways of doing this = 4 * 13C2
= 52 * 6.
5. When a
coin is tossed 10 times, what is the probability of getting more heads than
coin is tossed 'n' times, the probability of getting more heads than tails =
Probability of getting more tails than heads. So, we do not need to really
compute this. Probability = 1/2.
not counting the scenarios where there are an equal number of heads and tails.
If we threw the coin an odd number of times, the probability of getting more
heads than tails will be ½.
heads could be 6, 7, 8, 9, or 10. Probability = ( 10C6 + 10C7
+ 10C8 + 10C9 + 10C10)/
way of thinking about this would be ½ (1 – 10C5)/ 210.
If we removed the scenario where the number of heads and tails are equal, the
probability of more heads than tails should be equal to probability of more
tails than heads.
6. In how
many ways can we arrange 3 boys and 3 girls on a straight line such that no two
boys stand next to each other
arrange the 6 people as BGBGBG or GBGBGB. If we arranged them as BGBGBG, we
would have 3! * 3! ways of arranging the 3 boys and the 3 girls. So, total
number of possibilities = 2 * 3! * 3! = 72 ways.
not accounting for arrangement such as BGBGGB and BGGBGB. Note that the
question says that no two boys should stand next to each other. Two girls can
stand next to each other.
There are 4
possible outlines – BGBGBG, GBGBGB, BGBGGB or BGGBGB. So, total number of
possibilities = 3! * 3! * 4 = 144.
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