Question
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In a tennis tournament, the odds that player A will be the champion is 4 to 3 and the odds the player B will be the champion is 1 to 4. What are the odds that either A or B will become the champion.
(a) 25 to 8
(b) 27 to 8
(c) 4 to 1
(d) 27 to 35

Answer
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Hint: The information we are looking for is the probability of A becomes the champion or B which is the union of two events. This we find those 2 event’s probabilities separately and sum it to obtain the needed one.
* Odds a to b represent that the probability of champion ship is either $\dfrac{a}{{a + b}}$ or $\dfrac{b}{{a + b}}$ depending upon the person whose probability is to be found.
* Probability of an event =Number of favourable event/Total event
* Probability of either event1 or event2 = probability of event1 + probability of event2. Where these events are disjoint which means the probability of happening of event1 and event2 is zero.

Complete step-by-step answer:
Representing given information in probability form:
We use the formula $\dfrac{a}{{a + b}}$
Odds of A, 4 to 3 represents that a = 4 and b = 3 , so a+b = 4+3 = 7
Probability (A becomes the champion),
\[P(A) = \dfrac{4}{7}\]
Odds of B, 1 to 4 represents a = 1 and b = 4 , so a+b = 1+4 = 5
Probability (B becomes the champion),
\[P(B) = \dfrac{1}{5}\]
Probability that either A or B will become the champion
= P(A becomes champion) + P(B becomes champion.)
$P(A) + P(B) = \dfrac{4}{7} + \dfrac{1}{5}$
Take LCM on right hand side of the equation
$
  P(A) + P(B) = \dfrac{{4 \times 5 + 1 \times 7}}{{7 \times 5}} \\
  P(A) + P(B) = \dfrac{{27}}{{35}} \\
 $
These two events are disjoint. As happening in both events which are both A and B wins have probability zero.
Converting probability to odds.
Comparing the value \[\dfrac{{27}}{{35}}\] to $\dfrac{a}{{a + b}}$, we get \[a = {\text{ }}27,a + b = 35,\]
\[ \Rightarrow b = 35 - a \Rightarrow b = 35 - 27 = 8\]
So, a = 27 and b = 8
Odds that A or B will become the champion is 27 to 8

So, the correct answer is “Option b”.

Note: Students might make the mistake of calculating one of the probabilities and then subtracting the probability from 1 to get another probability which is wrong because we are given two separate odds here.