Maths~Probability (2 questions)

1)
n = 300
p = 0.49

Pr(X = 150) = 300C150 x 0.49^150 x 0.51^(300-150) = 0.04334656

2)
Pr(fourth flip is a head | 3 heads occurred in the 4 flips)
= Pr(fourth flip is a head and 3 heads occurred in the 4 flips)
/ Pr( 3 heads occurred in the 4 flips)
= [3 x (1/2)^4] / [4 x (1/2)^4] = 3/4

where
Pr(fourth flip is a head and 3 heads occurred in the 4 flips) = Pr{THHH or HTHH or HHTH} = 3 x (1/2)^4

Pr( 3 heads occurred in the 4 flips) = Pr{THHH or HTHH or HHTH or HHHT}= 4 x (1/2)^4

actually, it is a binomial distribution

where probability mass function is

f(x) = nCx (p^x) [(1-p)^(n-x)]

now n = 300 and p = 0.49

by calculating the cumumlative distribution function F(x), you can get the answer

or alternatively,

by central limit theorem,

sqrt(sample no.) * [(sample mean) - (population mean)] / (population variance)

is in standard normal distribution i.e. Normal (0, 1)

so, if sample no. = 300

population mean = np = 300 * 0.49 = 147

population variance = np(1-p) = 300 * 0.49 * 0.51 = 74.97

so, we can now have the following equation:

10 * sqrt(3) * [(sample mean) - 147] / 74.97 = 0

1. P(half of them are hens) = 300C150 * (0.49)^150 * (0.51)^150
= 0.0433465603482866....

2. Method 1
P(4th is head | 3 heads in 4 flips)
= P(4th is head AND 3 heads in 4 flips) / P(3 heads in 4 flips)
= 3C2 * (0.5)^3 * 0.5 / (4C3 * (0.5)^3 * (0.5))
= 0.75

Method 2
Count the number....
HHHT NOT the one we want
HHTH the one we want
HTHH the one we want
THHH the one we want
Therefore Probability = 3/4 = 0.75