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Automobile engines and transmissions are produced on assembly lines, and are inspected for defects after they come off their assembly lines. Those with defects are repaired. Let X represent the number of engines, and Y the number of transmissions that require repairs in a one-hour time interval. The joint probability mass function of X and Y is as follows: y x 0 1 2 3 0 0.13 0.10 0.07 0.03 1 0.12 0.16 0.08 0.04 2 0.02 0.06 0.08 0.04 3 0.01 0.02 0.02 0.02 Find the marginal probability mass function of X. Round the answers to two decimal places. pX(0) pX(1) pX(2) pX(3)

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Answer:

[tex]\begin\\\begin{tabular}{|l|l|l|l|l|l|}x&0&1&2&4&Sum\\P(x)&0.33&0.40&0.20&0.07&1.00\end[/tex]

Step-by-step explanation:

From the question we are given the table

[tex]\begin\\\begin{tabular}{|l|l|l|l|l|} &&y \\x &0&1&2&3\\0 &0.13&0.1&0.07&0.03 \\1&0.12&0.16&0.08&0.04\\ 2&0.02&0.06&0.08&0.04\\ 3&0.01&0.02&0.02&0.02\end[/tex]

Generally the marginal probability mass function of X is mathematically given as

[tex]P(x)=\sum P(x,y)>y[/tex]

[tex]\sum P(0)=0.13+0.1+0.07+0.03=0.33\\\sum P(1)=0.12+0.16+0.08+0.04=0.40\\\sum P(2)=0.02+0.06+0.08+0.04=0.20\\\sum P(3)=0.01+0.02+0.02+0.02=0.07[/tex]

Therefore marginal probability mass function of X

[tex]\begin\\\begin{tabular}{|l|l|l|l|l|l|}x&0&1&2&4&Sum\\P(x)&0.33&0.40&0.20&0.07&1.00\end[/tex]

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