# Three coin flips - two biased and one fair. Is it a Binomial distribution?

Suppose one coin with probability $$0.4$$ for heads, one with probability $$0.6$$ for heads, and one that is a fair coin are each tossed once. Find the pmf of the total number of heads obtained. Is it a Binomial distribution?

Here's what I am thinking:

$$X=$$ # of heads

When $$X=0$$

No head $$(TTT)$$ which means $$0.6$$ for the first one, $$0.4$$ for the second one and $$0.5$$ for the third one. Therefore, $$P(X=0) = (0.6)(0.4)(0.5) = 0.12$$

Similarly,it will be $$P(X=3) = (0.6)(0.4)(0.5) = 0.12$$

I am stuck when $$X=1$$ and $$X=2$$

When $$X=1$$ which means it will be $$\left \{ HTT,THT,TTH \right \}$$

Let's say the first coin always is the one with Heads. So, we have $$3*(0.4)(0.4)(0.5)$$

And when the Head is from the second coin, we have $$3*(0.6)(0.6)(0.5)$$

And when it's coin three, we have $$3*(0.6)(0.4)(0.5)$$

So, $$P(X=1)$$ will be sum of all these. Same thing for $$P(X=2)$$

I am not seeing a binomial distribution because $$P(X=1)$$ is giving me after adding all the terms above $$3*(0.5)\left \{(0.4)^2 + (0.6)^2 + (0.6)(0.4) \right \}$$

Am I doing it right?

• For $P(X=1)$, why multiply the probability of each individual outcome by $3$? Jan 12, 2020 at 18:24
• because there are three different outcomes {HTT,THT,TTH} ? Jan 12, 2020 at 18:27
• have a look at my answer below. Jan 12, 2020 at 18:47

Let $$X$$ denote the number of heads obtained when these three coins are tossed once each. Then the possible valuse for $$X$$ are $$x = 0, 1, 2, 3$$.
We find that $$P(X=0) = P\big( \{TTT \} \big) = (1-0.4)(1-0.6)(1-0.5) = (0.6)(0.4)(0.5) = 0.12,$$ \begin{align} P(X = 1) &= P\big(\{ HTH, THT, TTH \} \big) = P \big(\{ HTT \}\big) + P \big( \{ THT \} \big) + P \big( \{ TTH \} \big) \\ &= (0.4)(1-0.6)(1-0.5) + (1-0.4)(0.6)(1-0.5) + (1-0.4)(1-0.6)(0.5) \\ &= (0.4)(0.4)(0.5) + (0.6)(0.6)(0.5) + (0.6)(0.4)(0.5) \\ &= 0.08 + 0.18 + 0.12 \\ &= 0.38, \end{align} \begin{align} P(X = 2) &= P\big(\{ HHT, HTH, THH \} \big) = P \big(\{ HHT \}\big) + P \big( \{ HTH \} \big) + P \big( \{ THH \} \big) \\ &= (0.4)(0.6)(1-0.5) + (0.4)(1-0.6)(0.5) + (1-0.4)(0.6)(0.5) \\ &= (0.4)(0.6)(0.5) + (0.4)(0.4)(0.5) + (0.6)(0.6)(0.5) \\ &= 0.12 + 0.08 + 0.18 \\ &= 0.38, \end{align} and finally $$P(X=3) = P\big( \{ HHH \} \big) = (0.4)(0.6)( 0.5) = 0.12.$$
In order to double check, we find that $$P(X = 0) + P(X = 1) + P(X = 2) + P(X=3) = 0.12 + 0.38 + 0.38 + 0.12 = 1.$$