Probability

Question

an unfair coin has a probability of coming up heads of 0.65. The coin is flipped 50 times. What is the probability it will come up heads 25 or fewer times? (Give answer to at least 3 decimal places). (relevant section)

Answer 1

This is just P(X ≤ 25) where X ~ Bin(n=50,p=0.65)

 

Using R by typing pbinom(25,size=50,prob=0.65), we have:

 

P(X ≤ 25) = 0.02066839

 

We can also give a normal approximation

 

Let Y ∼ N(μ,σ²) where

 

μ = np = 32.5

 

σ = √[np(1-p)] = 3.37268

 

Then, using continuity correction, P(X ≤ x) ≈ P(Y ≤ x + 1/2) gives

 

Z = (Y - μ)/σ = (X - μ + 1/2)/σ =(25 - 32.5 + 0.5)/3.37268 = -2.075498

 

P(X ≤ 25) ≈ P(Z ≤ -2.075498) = Φ(-2.075498) ≈ 0.0189702

 

this is close to the exact value. The command here is pnorm(-2.075498).

Answer 2

This is an example of binomial (Bernoulli) distribution.

 

If we let p = 0.35 (probability of TAILS outcome on one 'trial'), and n = 50 is the number of trials, then the TI-83/84 DISTR command  BINOMIALCDF(50,0.35,25) will give the cumulative value of 26 terms, the first of which represents the probability of 0 'Tails' /50 'Heads' in a 50-trial experiment, the second represents the probability of 1 'Tails'/49 'Heads', etc. up to the last of 25 terms, which represents 25 Tails/25 Heads.

 

This value is approximately 0.989956. Rounded to 3 DP: 0.990 ... it's the probability that the experiment comes up heads 50 or 49 or ...25 times, i.e.  P(0 to 25 Tails).

 

To compute the probability of 25 Heads or LESS, you calculate 1 - P(0 to 25 Tails) = 0.010

 

NOTE; At first I tried binomcdf(50,0.65,25) but that did not work (possibly 'overflow' problem within the calculator for such a combination of large n and p > 0.5), so the above was my workaround solution.

 

I WILL WELCOME ANY COMMENTS FROM MY FELLOW TUTORS.  Thanks.