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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?

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

2 Answers 2

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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 $$ P(X = 0) + P(X = 1) + P(X = 2) + P(X=3) = 0.12 + 0.38 + 0.38 + 0.12 = 1. $$

In short, this distribution is not exactly the binomial distribution.

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It isn't a binomial distribution, because the probability of success/failure in each outcome isn't the same. You will have to do it the old-fashioned way.

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