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I was doing some random problems on statistics after covering few topics till Moments. The following question puzzled me. I am not able to put any theory.

Tom and Joy tied for the first place in a debate competition. The winner is to be decided by the majority opinion of a panel of three judges chosen at random from a group of seven judges. If four of these judges favour Tom and three favour Joy, what is the probability that Tom will be declared the winner.

Kindly help me to solve this problem. Is there any specific topic to find solutions of these kinds of problems?

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    Are you familiar with the Binomial Coefficient? (Topic: Selection and Combinations) – 2017-02-15

2 Answers 2

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The topic is: Selection and Combinations.

When outcomes of the sample space are all of equal probability weight, then the probability of an event is just the ratio of size of the event to the size of the sample space.

Are you familiar with the use of the binomial coefficient to count selection of items from a set?

You seek the probability for selecting two or three from the four judges who favour Tom, and none or one (respectively) from the three who favour Joy, when selecting any three from all seven judges.

$$\dfrac{\dbinom 42\dbinom 31 + \dbinom 43\dbinom 30}{\dbinom 73}~=~\dfrac{22}{35}~=~0.6\dot{\overline{285714}}$$

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    Yeah I just read after reading your answer. Thank you so much for the help. – 2017-02-15
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Edit 1: The previous answer was wrong (partially)

Total possible choices of judges which makes Tom a winner is $\binom{4}{2}\binom{3}{1} + \binom{4}{3}\binom{3}{0}$. Thus @George's answer holds.

Similar kinds of problems are solved with the knowledge of https://en.wikipedia.org/wiki/Binomial_coefficient

The previous answer (wrong one):

Total possible choices of judges is $7 \choose 3$. Total possible choices of judges which makes Tom a winner is $4 \choose 2$$\cdot 5.$ The quantity you are interested in is $$\frac{5 \binom{4}{2}}{7 \choose 3} = \frac{6}{7}.$$

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    You are over counting the cases for selecting three judges who favour Tom. – 2017-02-15
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    Thanks Mr. @GrahamKemp for correcting me. – 2017-02-15
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    @GrahamKemp The over counting which I did earlier was true if the identity of the judges were unknown right? "Like in permutation" – 2017-02-15
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    You counted ways to select two from four favouring judges then one from the remaining five judges - however that third judge could be another favouring judge and order of selection does not matter. Selecting A,B then C is the same outcome as selecting A,C then B. – 2017-02-15