Question

Let X be a discrete random variable for the outcome of a toss of a coin with sides 0 and 1,with P(X = 0) = 0.5 and P(X = 1) = 0.5. Moreover let Y be another variable associatedwith a biased coin with sides 1 and 2, such that P(Y = 1) = 0.4 and P(Y = 2) = 0.6. IfZ = X + Y , what is P(Z < 3)

Answer 1

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given probabilities

`\begin{gathered} P(X=0)=0.5,P(X=1)=0.5 \\ P(Y=1)=0.4,P(Y=2)=0.6 \\ Z=X+Y \end{gathered}`

STEP 2: Write the formula for calculating the required probability

`P(Z<3)=P(Z=1)+P(Z=2)`

STEP 3: Find P(Z=1)

`\begin{gathered} P(Z=1)=P(X=0)\text{ and }P(Y=1) \\ =0.5*0.4=0.2 \end{gathered}`

STEP 4: Find P(Z=2)

`\begin{gathered} P(Z=2)=P(X=0)\cdot P(Y=2)\text{ or }* P(X=1)\cdot P(Y=1) \\ (0.5*0.6)+(0.5*0.4) \\ =0.3+0.2=0.5 \end{gathered}`

STEP 5: Find the P(Z<3)

`P(Z<3)=0.5+0.2=0.7`

Hence, the answer is 0.7