Probability
Aptitude

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Q.

The odds against an event are 5:3 and the odds in favour of another independent event are 7:5. Find the probability that at least one of the two events will occur.

 A.

52/96

 B.

69/96

 C.

71/96

 D.

13/96

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Solution:
Option(C) is correct

Let probability of the first event taking place be $A$ and probability of the second event taking place be $B$.

Then,

$\begin{align*}
P(A)&= \dfrac{3}{5+3}\\
&= 3/8\\
P(B) &= \dfrac{7}{7+5}\\
&= 7/12
\end{align*}$

The required event can be defined as that $A$ takes place and $B$ does not take place ($A$ or $B$ takes place and $A$ does not take place or $A$ takes place and $B$ takes place.)

$= [P(A)\{1-P(B) \}] + [P(B)\{1-P(A)\}] + [P(A)P(B)]$

$= \left[\dfrac{3}{8} × \dfrac{5}{12}\right]+\left[\dfrac{5}{8} × \dfrac{7}{12}\right]+\left[\dfrac{3}{8} × \dfrac{7}{12}\right]$

$= \dfrac{15+35+21}{96}$

$= \dfrac{\textbf{71}}{\textbf{96}}$


(9) Comment(s)


Anonymous
 ()

The calculation of the Probability P(B) based on the the odds is not correct.

P(B) should be 5/12 instead of 7/12.

P(A)=3/8

The final result should be P(A U B)=61/96.

The different option to obtain this result are :

3/8+5/12-3/8*5/12=61/96

or

1-(5/8*7/12)=61/96

or

(5/8*5/12)+(3/8*7/12)+(3/8*5/12)=25/96+21/96+15/96=61/96



Bhagyashree
 ()

P(A)=3/(3+5). I.e. odd against

P(B)=7/(7+5). I.e. odd favor

At least one of the event occur, we are interested in finding P(A U B)

So we know,

P(A U B)= 1- P(!A)P(!B)

1- (5/8)(5/12)= 71/96



Harish Kumar
 ()

1 - none of them occur =1-(5/12 * 5/8)=71/96



Dileep Barnwal
 ()

A-occurrence against

$P(A)=3/8$, $P(A')=5/8$

B-occurrence in favour

$P(B)=7/12$, $P(B')=5/12$

Neither $A$ nor $B$ occur is $P(A')*P(B')=25/96$

Alteast one of them occurs is $1-25/96=71/96$



Niranjana Murali
 ()

Hi all,

I couldn't understand how $P(A)$ and $P(B)$ are calculated?

Can anyone explain?



Mohini
 ()

Hi,

according to me, if we r taking

in the 1st event: against: 3/8, favor- 5/8

in the 2nd event: Against : 5/12, Favor 7/12

There r three chance for at least one event occur: (1st against*2nd favor)+(1st favor* 2nd against)+(both in favor).

thus: (3/8*7/12)+(5/8*5/12)+(5/8*7/12)

21/96 +25/96 + 35/96 = 81/ 96

Kindly suggest if I am wrong and how?


Reddy
 ()

Smile$5/8+7/12-35/96=81/96$

Shaswat Khamari
 ()

Hi,
according to me, if we r taking
in the 1st event: against: 3/8, favor- 5/8
in the 2nd event: Against : 5/12, Favor 7/12

There r three chance for at least one event occur: (1st against*2nd favor)+(1st favor* 2nd against)+(both in favor).
thus: (5/8*7/12)+(3/8*5/12)+(3/8*7/12)
35/96 +15/96 + 21/96 = 71/ 96


Kiran
 ()

it is very good