If the mode is 10, then both P(X = 9) and P(X = 11) must be smaller than P(X = 10)

i.e P(X = 9) < P(X = 10) and P(X = 11) < P(X = 10)


Put in the formula for finding the probability, (n x) pˣ (1 - p)ⁿ⁻ˣ

(24 9) p⁹ (1 - p)²⁴⁻⁹ < (24 10) p¹⁰(1 - p)²⁴⁻¹⁰

24!/15!9! p⁹ (1 - p)¹⁵ < 24!/14!10! p¹⁰ (1 - p)¹⁴

Dividing both sides by 24! and cross multiplying the other factorials,


10p⁹ (1 - p)¹⁵ < 15p¹⁰ (1 - p)¹⁴

2p⁹ (1 - p)¹⁵ < 3p¹⁰ (1 - p)¹⁴

Since p is a constant, p > 0 and 1 - p > 0 , we can divide both sides by p⁹ and (1 - p)¹⁴

(We can't have negative probability (p ≮ 0) and p ≠ 0 since it would mean P(X = x) = 0 for any x from 0 to 24)

2(1 - p) < 3p

2 - 2p < 3p

2 < 5p

2/5 < p

wow ok thanks much!

Likewise for P(X = 11) < P(X = 10) ,


(24 11) p¹¹ (1 - p)²⁴⁻¹¹ < (24 10) p¹⁰ (1 - p)²⁴⁻¹⁰

24!/13!11! p¹¹ (1 - p)¹³ < 24!/14!10! p¹⁰ (1 - p)¹⁴


Dividing both sides by 24! and cross multiplying the other factorials,


14 p¹¹ (1 - p)¹³ < 11 p¹⁰ (1 - p)¹⁴

Since p is a constant, p > 0 and 1 - p > 0 , we can divide both sides by p¹⁰ and (1 - p)¹³


14p < 11(1 - p)

14p < 11 - 11p

25p < 11

p < 11/25

So since p > 2/5 (0.4) and p < 11/25 (0.44),

Overall →2/5 < p < 11/25

Edited for errors.