Flipping a Biased Coin

Probability Level 3

There is a biased coin which has a probability $\frac{3}{8}$ of being heads and probability $\frac{5}{8}$ of being tails.

What is the number of times $n$ where $n≥2$ that it should be flipped such that the probability of getting two heads is equal to the probability of getting three heads?

The answer is 7.

1 solution

Nick Turtle
Oct 29, 2017

The probability of getting $2$ heads and $n-2$ tails, in that order, is ${(\frac{3}{8})}^2{(\frac{5}{8})}^{n-2}$ . There are $\frac{n!}{2!(n-2)!}$ way of ordering it. So, the probability of getting $2$ heads and $n-2$ tails, in any order, is $\frac{n!}{2!(n-2)!}{(\frac{3}{8})}^2{(\frac{5}{8})}^{n-2}$ .

Do the same for $3$ heads and $n-3$ tails, which results to $\frac{n!}{3!(n-3)!}{(\frac{3}{8})}^3{(\frac{5}{8})}^{n-3}$ .

These two values are equal, so we solve for $n$ in $\frac{n!}{2!(n-2)!}{(\frac{3}{8})}^2{(\frac{5}{8})}^{n-2}=\frac{n!}{3!(n-3)!}{(\frac{3}{8})}^3{(\frac{5}{8})}^{n-3}$ .

Many terms cancel out, and we are left with $3\cdot\frac{5}{8}=\frac{3}{8}(n-2)$ . Thus, we have $n=7$ .

# There is a biased coin which has a probability 3/8 of being heads and probability 
# 5/8 of being tails.
# What is the number of times n where n >= 2  that it should be flipped such 
# that the probability of getting two heads is equal to the probability of getting three heads?

p_heads = float(3/8.)
p_tails = 1-p_heads

heads_1 = 2
heads_2 = 3
num_of_flips = ((heads_2*p_tails)/p_heads)+heads_1
print num_of_flips

Michael Fitzgerald - 3 years, 7 months ago

Flipping a Biased Coin

The answer is 7.

1 solution

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