Probabilities 71194

We have a dummy die where numbers fall with probabilities P (1)=0.1; P (2)=0.2; P (3)=0.22; P (4)=0.16; P (5)=0.24; P (6)=0.08. Determine the probability that the two toss the same numbers.

Correct answer:

p =  0.188

Step-by-step explanation:

$$\begin{gathered} p_1=0.1 \\ {p}_2=0.2 \\ {p}_3=0.22 \\ {p}_4=0.16 \\ {p}_5=0.24 \\ {p}_6=0.08 \\ p=p_1^2+p_2^2+p_3^2+p_4^2+p_5^2+p_6^2=0.1^2+0.2^2+0.22^2+0.16^2+0.24^2+0.08^2=\frac{47}{250}=0.188 \\ \text{ Verifying Solution: } \\ {p}_0=\frac{1}{6}\doteq 0.1667 \\ {p}_8=6\cdot p_0^2=6\cdot {\frac{1}{6}}^2=\frac{1}{6}\doteq 0.1667 \end{gathered}$$

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