Probability
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Q.

An urn contains 6 red, 5 blue and 2 green marbles. If 2 marbles are picked at random, what is the probability that both are red?

 A.

6/13

 B.

5/26

 C.

5/13

 D.

7/26

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Solution:
Option(B) is correct

P(Both are red),

$= \dfrac{^6C_2}{^{13}C_2}$

$= \textbf{5/26}$

Edit: For an alternative solution, check comment by Vejayanantham TR.


(5) Comment(s)


Bishnu Chetri
 ()

Let A=event of selecting a red marble on first draw

B=event of selecting a red marble on the second draw

According to stochastic process,

P(A intersection B)=P(A).P(B|A)

=(6/13).(5/12)

=5/6



Vejayanantham TR
 ()

Another way,

Red balls = 6,

In first take , probability to take a red one is = 6/13

Now ,there is only 5 red ball,

Probability to take a red one = 5/12

Total probability = 6/13 * 5/12 = 5/26



Dhamu
 ()

Why took $\dfrac{5}{26}$ there is no relation for $\dfrac{5}{26}$.

Please explain to reach my knowledge clearly



Aarti
 ()

How you have arrived at $\dfrac{5}{26}$

and $\dfrac{^6c_2}{^{13}c_2} i am not unerstandingf please teach me in simple words.


Bharath
 ()

6C2= 6*5/2*1, 13c2= 13*12/2*1....

Generally, XCY= X*X-1*X-2.../Y*Y-1*Y-2...

If x = 14 and y = 3 then in numerator we need to multiply three numbers (=value of Y), means 14*13*12 and in denominator 3*2*1