Probability
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Q.

In a race, the odd favour of cars $P,Q,R,S$ are $1:3, 1:4, 1:5$ and $1:6$ respectively. Find the probability that one of them wins the race.

 A.

9/17

 B.

114/121

 C.

319/420

 D.

27/111

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Solution:
Option(C) is correct

Let the probability of winning the race is denoted by $P(\text{person})$

$\(P(P)=\dfrac{1}{4},\;P(Q)=\dfrac{1}{5},\;P(R)=\dfrac{1}{6},\;P(S)=\dfrac{1}{7}\)

All the events are mutually exclusive (since if one of them wins then other would lose as pointed out by rahul) hence,

Required probability: 

\(=P(P)+P(Q)+P(R)+P(S) \)

\(=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}\)

\(=\dfrac{319}{420}\)


(11) Comment(s)


Anonymous
 ()

The probability of winning is not correct.

It should be 3/4+4/5+5/6+6/7

If odds are stated as an A : B chance of success then the probability of success is given as P = B / (A + B)



Shristi
 ()

if one wins then others are definitely losing then we have to multiply their losing probability too



Payal
 ()

the probabilities taken for calculation are for their failure. I guess we need to use 1-1/4...etc for all to make sure if one wins


Payal
 ()

it should be 3/4+4/5+5/6+6/7


Poonam Pipaliya
 ()

can't we do like this,

$P(A)(1-P(B))(1-P(C))$

because here one wins then others definitely loose.


Miya
 ()

That is only one of the possibilities, what about $P(B)(1-P(A))(1-P(C))$ or similar events where $B$ wins?

One may proceed with your approach but that would take lot of counting for larger numbers.

It would be better to take advantage of counting techniques such as used in solution.


Rahul
 ()

how are the events mutually exclusive ?

If P wins the race then others will lose the same race.


Deepak
 ()

Yes, Rahul, you answered it yourself.

Based on your input, updated the solution.

Thank you for for making it more useful for others.


Sumit
 ()

I think the answer is wrong.



Abhishek
 ()

kindly explain, how come it is mutually exclusive....



Vikas
 ()

Question has different values than answer values!