Three Toronto Maple Leaf fans attend a Flames-Leafs game in the Saddledome. The probability that the first fan will wear their "Leafs" jersey is 0.79. The probability that the second fan will wear their "Leafs" jersey is 0.54. The probability that the third fan will not wear their "Leafs" jersey is 0.79. Let X be a random variable which measures how many of the three Leaf fans mentioned are wearing their "Leafs" jersey to this hockey game. Assuming that each "Leaf" fan mentioned wears their "Leaf" jersey independently of each other,find the probability distribution of X.

find P(X=0), p(X=1). P(X=2), P(X=3)

I know P(X=3) = 0.79(0.54)(1-0.79) and therefore P(x=0) = 1-P(x-3) but I don't know how to find P(x=1) and P(X=2)

Also, what kind of distribution is this?



1 Answer 1


Since the probability of each fan wearing their Leafs jerseys is different, this isn't any specific kind of distribution. (If the three probabilities were identical, it would be a binomial distribution.) Also, it is not the case that $P(X = 0) = 1-P(X = 3)$; that would be the case only if it were impossible for $X = 1$ or $2$.

Let's put everyone on an equal footing, so that $p_1 = 0.79$, $p_2 = 0.54$, and $p_3 = 0.21$ are the probabilities of wearing the jersey. Then

$$ P(X = 0) = (1-p_1)(1-p_2)(1-p_3) $$

$$ P(X = 1) = p_1(1-p_2)(1-p_3) + (1-p_1)p_2(1-p_3) + (1-p_1)(1-p_2)p_3 $$

$$ P(X = 2) = p_1p_2(1-p_3) + p_1(1-p_2)p_3 + (1-p_1)p_2p_3 $$

$$ P(X = 3) = p_1p_2p_3 $$

In each case, we have to enumerate the different ways that we can obtain a particular value for $X$.

Note that there are $1$, $3$, $3$, and $1$ different ways to obtain $X = 0$, $1$, $2$, and $3$, so if $p_i = p$ for all $i$, then we could write

$$ P(X = k) = \binom{3}{k} p^k(1-p)^{3-k} $$

which is the binomial distribution for population size $N = 3$ and success probability $p$.


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