From the table, we have:
$P(X=0, Y=0) = 0.31$
$P(X=0, Y=1) = p$
$P(X=0, Y=2) = 0.05$
$P(X=1, Y=0) = 0.21$
$P(X=1, Y=1) = 0.13$
$P(X=1, Y=2) = q$
Since the sum of all probabilities must be 1, we have:
$0.31 + p + 0.05 + 0.21 + 0.13 + q = 1$
$0.7 + p + q = 1$
$p +
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