UPSC NDA I 2024 Mathematics Solved Paper

To determine the value of p+q, let's first analyze the given probabilities for an unbiased coin being tossed n times.
The probability of getting at least one tail when an unbiased coin is tossed n times is given by:
p=1− Probability of getting all heads
Since the probability of getting all heads (which is HHH...H ) on n coin tosses is (‌

1

2

)n, we have:
p=1−(‌

1

2

)n
The probability of getting at least two tails is:
q=1− Probability of zero tails - Probability of one tail
The probability of zero tails is (‌

1

2

)n, and the probability of exactly one tail (which can occur in any of the n positions) is n⋅(‌

1

2

)⋅(‌

1

2

)n−1=n⋅(‌

1

2

)n. Therefore,
q=1−(‌

1

2

)n−n⋅(‌

1

2

)n
We are given that:
p−q=‌

5

32

Substituting the expressions for p and q, we get:
(1−(‌

1

2

)n)−(1−(‌

1

2

)n−n⋅(‌

1

2

)n)=‌

5

32

Simplifying the equation above:
1−(‌

1

2

)n−1+(‌

1

2

)n+n⋅(‌

1

2

)n=‌

5

32

n⋅(‌

1

2

)n=‌

5

32

We need to solve for n :
Let's say n=5, then:
5⋅(‌

1

2

)5=‌

5

32

Which simplifies to:
‌

5

32

=‌

5

32

This shows that our assumption n=5 is correct. Now, substituting n=5 back into the expressions for p and q :
p=1−(‌

1

2

)5=1−‌

1

32

=‌

31

32

q=1−(‌

1

2

)5−5⋅(‌

1

2

)5=1−‌

1

32

−‌

5

32

=1−‌

6

32

=‌

26

32

Adding p and q :
p+q=‌

31

32

+‌

26

32

=‌

57

32