A bag contains 10 balls out of which k are red and (10−k) are black, where 0≤k≤10. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:
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Official solution
Given Terms
Total Balls ( n ) : 10
Red Balls ( k ) : Variable from 0 to 10 .
Black Balls : (10−k).
Sample Drawn ( r ) : 3 balls, without replacement.
Observed Event (A) : All 3 balls drawn are black.
Target Event : Find the probability that the bag contained exactly 1 red and 9 black balls (which means k=1 ).
Let Ek be the event that the bag contains k red balls. Since k can be any integer from 0 to 10 , there are 11 possible compositions. Assuming they are equally likely:
P(Ek)=111
The event A is drawing 3 black balls. The probability of this happening depends on how many black balls (10−k) were in the bag :
P(A∣Ek)=10C310−kC3
Note : For this to be possible, there must be at least 3 black balls, so 10−k≥3, meaning k≤7.
Using Bayes' Theorem
We need P(E1∣A), which is the probability of having 1 red ball given that 3 black balls were drawn :
P(E1∣A)=∑k=07P(Ek)⋅P(A∣Ek)P(E1)⋅P(A∣E1)
Since P(Ek) is constant (111) for all k, it cancels out :
P(E1∣A)=10C3+9C3+⋯+3C310−1C3
denominator 10C3+9C3+⋯+3C3 can be written as 3C3+4C3+5C3⋯10C3
using identity kCk+k+1Ck+k+2Ck+…nCk=n+1Ck+1
so, denominator become =11C4=4×3×2×111×10×9×8=330
Numerator : 10−1C3=9C3=3×2×19×8×7=84P(E1∣A)=33084=5514