A bag contain 5 blue and 4 black balls. Three balls are drawn at random. What is the probability tha

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13 Nov 2015 Probability

A bag contain 5 blue and 4 black balls. Three balls are drawn at random. What is the probability that 2 are blue and 1 is black.

OPtion

1) 3/5
2) 2/7
3) 1/3
4) 1/5
5) 1/4
6) 2/3
7) 2/5
8) 1/6
9) 1/7
10)None of these

Solution

n(s)=9C3
n(E)=(5C2*4C1)

p(E)=n(E)/n(S)
=(5C2*4C1)/9C3
=10/21

10/21

Option 10)

Correct Option: 10 Best Solution (6)

Jaikumar Bhatia   9 years AGO

n(s)=9C3
n(E)=(5C2*4C1)

p(E)=n(E)/n(S)
=(5C2*4C1)/9C3
=10/21

hence ans is 10/21
ie option is (10)..none of these

Like? Yes (5) | No   

Anusha   9 years AGO

Favorable is 2 blue balls and 1 black ball
Possible is 3 blue balls 0 black ball,2 blue balls 1 black ball,1 blue ball 2 black balls,0 blue balls 3 black balls
So, (5C2*4C1)/(5C3*4C0+5C2*4C1+5C1*4C2+5C0*4C3)
=(10*4)/(10*1+10*4+5*6+1*4)
=40/84
=10/21
So, the option is none of these 10)None of these

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Vivek Singh   9 years AGO

n(S)= 9*8*7/3*2*1 = 84
5*4/2*1= 10 and 4/1=4 =>
n(E)=10*4=40
probablity= n(E)/n(S) = 40/84= 10/21 so none of these

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MURUGESAN.P   9 years AGO

selecting two blue ball out of 5 blue ball:5C2=10
selecting 1 black ball out of 4 black black ball:4c1=4
therefore
no of ways=5C2 * 4C1=40
total no of ways=9C3=84
probability=40/84 (ie) 10/21
option 10:none of these

Like? Yes (3) | No   

chetan   9 years AGO

Probability=(5c2*4c1)/9c3=10/21.
So option 10 is correct.

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Nandkishor jaiswal   9 years AGO

total number of events will be 9c3=9*8*7/3!=84
posibility for choosing 2 blue out of 5 is 5c2=5*4/2=10
posibility for choosing 1 black out of 4 is 4c1=4
so total probability is 10*4/84=10/21

Like? Yes (3) | No