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From Wednesday through Saturday, a library subscribes to two separate weekly news publications, each of which has a high probability of arriving late.

Each individual probability is given as:

\(\begin{array}{l}P\left( {{\rm{Wed}}} \right) = 0.30\\P\left( {{\rm{Thurs}}} \right) = 0.40\\P\left( {{\rm{Fri}}} \right) = 0.20\\P\left( {{\rm{Sat}}} \right) = 0.10\end{array}\)

Each magazine can arrive between Wednesday to Saturday. So, the possible number of outcomes are\(\left\{ {{\rm{Wed,Thurs,Fri,Sat}}} \right\}\).

If there are two magazines, then the possible number of outcomes is\({4^2} = 16\).

Let denote Wed as W, Thurs as T, Fri as F and Sat as S. The sample space S is given as:

\(S = \left\{ \begin{array}{l}\left( {W,W} \right),\left( {W,T} \right),\left( {W,F} \right),\left( {W,S} \right)\\\left( {T,W} \right),\left( {T,T} \right),\left( {T,F} \right),\left( {T,S} \right)\\\left( {F,W} \right),\left( {F,T} \right),\left( {F,F} \right),\left( {F,S} \right)\\\left( {S,W} \right),\left( {S,T} \right),\left( {S,F} \right),\left( {S,S} \right)\end{array} \right\}\)

Consider Y to be the number of days beyond Wednesday that it takes for both magazines to arrive.

The value associated with Y are as follows:

\(\begin{array}{l}Y\left( {W,W} \right) = 0\\Y\left( {W,T} \right) = 1\\Y\left( {W,F} \right) = 2\\Y\left( {W,S} \right) = 3\end{array}\)

\(\begin{array}{l}Y\left( {T,W} \right) = 1\\Y\left( {T,T} \right) = 1\\Y\left( {T,F} \right) = 2\\Y\left( {T,S} \right) = 3\end{array}\)

\(\begin{array}{l}Y\left( {F,W} \right) = 2\\Y\left( {F,T} \right) = 2\\Y\left( {F,F} \right) = 2\\Y\left( {F,S} \right) = 3\end{array}\)

\(\begin{array}{l}Y\left( {S,W} \right) = 3\\Y\left( {S,T} \right) = 3\\Y\left( {S,F} \right) = 3\\Y\left( {S,S} \right) = 3\end{array}\)

From the above list of values, the associated values of random variable Y are\(\left\{ {0,1,2,3} \right\}\).

As Y is a random variable that the number of days beyond Wednesday that it takes for both magazines to arrive, when\(Y = 0\) it means the maximum number of days is 0.

The possible pair is \(\left\{ {\left( {W,W} \right)} \right\}\).

So, its probability can be calculated as:

\(\begin{aligned}P\left( {Y = 0} \right) &= P\left( {WW} \right)\\ &= 0.30 \times 0.30\\ &= 0.09\end{aligned}\)

Similarly, when\(Y = 1\) it means the maximum number of days is 1.

The possible pairs are\(\left\{ {\left( {W,T} \right),\left( {T,W} \right),\left( {T,T} \right)} \right\}\).

So, its probability is calculated as:

\(\begin{aligned}P\left( {Y = 1} \right) &= \left( {P\left( {W,T} \right) + P\left( {T,W} \right) + P\left( {T,T} \right)} \right)\\ &= \left( {\left( {0.30} \right)\left( {0.40} \right) + \left( {0.30} \right)\left( {0.40} \right) + \left( {0.40} \right)\left( {0.40} \right)} \right)\\ &= \left( {0.12 + 0.12 + 0.16} \right)\\ &= 0.40\end{aligned}\)

Similarly, when\(Y = 2\) it means the maximum number of days is 2.

The possible pairs are\(\left\{ {\left( {W,F} \right),\left( {F,W} \right),\left( {T,F} \right),\left( {F,T} \right),\left( {F,F} \right)} \right\}\).

So, its probability is calculated as:

\(\begin{aligned}P\left( {Y = 2} \right) &= \left( {P\left( {W,F} \right) + P\left( {F,W} \right) + P\left( {T,F} \right) + P\left( {F,T} \right) + P\left( {F,F} \right)} \right)\\ &= \left( \begin{array}{l}\left( {0.30} \right)\left( {0.20} \right) + \left( {0.20} \right)\left( {0.30} \right) + \left( {0.40} \right)\left( {0.20} \right)\\ + \left( {0.20} \right)\left( {0.40} \right) + \left( {0.20} \right)\left( {0.40} \right)\end{array} \right)\\ &= \left( {0.06 + 0.06 + 0.08 + 0.08 + 0.08} \right)\\ &= 0.32\end{aligned}\)

Finally, when\(Y = 3\) it means the maximum number of days is 3.

The possible pairs are\(\left\{ {\left( {W,S} \right),\left( {T,S} \right),\left( {S,S} \right),\left( {F,S} \right),\left( {S,W} \right),\left( {S,F} \right),\left( {S,T} \right)} \right\}\).

So, its probability is calculated as:

\(\begin{aligned}P\left( {Y = 3} \right) &= \left( \begin{array}{l}P\left( {W,S} \right) + P\left( {T,S} \right) + P\left( {S,S} \right) + P\left( {F,S} \right)\\ + P\left( {S,W} \right) + P\left( {S,F} \right) + P\left( {S,T} \right)\end{array} \right)\\ &= \left( \begin{array}{l}\left( {0.30} \right)\left( {0.10} \right) + \left( {0.40} \right)\left( {0.10} \right) + \left( {0.10} \right)\left( {0.10} \right) + \left( {0.20} \right)\left( {0.10} \right)\\ + \left( {0.10} \right)\left( {0.30} \right) + \left( {0.10} \right)\left( {0.20} \right) + \left( {0.10} \right)\left( {0.40} \right)\end{array} \right)\\ &= \left( {0.03 + 0.04 + 0.01 + 0.02 + 0.03 + 0.02 + 0.04} \right)\\ &= 0.19\end{aligned}\)

The probability mass function of Y is given as: