UPSC NDA I 2025 Mathematics Paper

Given,
Let X be a random variable following a binomial distribution with parameters n=6 and p=k.
Further, it is given that:
9P(X=4)=P(X=2)
The probability mass function for a binomial distribution is:
P(X=x)=(

n

x

)px(1−p)n−x
For P(X=4), we have:
P(X=4)=(

6

4

)k4(1−k)2=15k4(1−k)2
For P(X=2), we have:
P(X=2)=(

6

2

)k2(1−k)4=15k2(1−k)4
We are given that:
9P(X=4)=P(X=2)
Substitute the expressions for P(X=4) and P(X=2) :
9×15k4(1−k)2=15k2(1−k)4
Cancel the common factor of 15 :

‌9k4(1−k)2=k2(1−k)4
‌9k2=(1−k)2
‌9k2=1−2k+k2
‌8k2+2k−1=0
Use the quadratic formula to solve for k :
‌k=‌

−2±√22−4×8×(−1)

2×8

‌k=‌

−2±√4+32

16

‌k=‌

−2±√36

16

‌k=‌

−2±6

16

Thus, the two possible values for k are:
k=‌

4

16

=‌

1

4

‌ or ‌k=‌

−8

16

=−‌

1

2

Since k represents a probability, it must be between 0 and 1 . Therefore, the valid solution is:
k=‌

1

4