Applied Mathematics for Electrical Engineering (3130908)

Question-4c ( ii )

BE | Semester-3 Winter-2019 | 26-11-2019

Q4) (c ( ii ))

No. of car having tyre problem (X) = 2.

No. of car having break problem (Y) = 1.

No. of car which are in good running condition = 2.

Therefore, the joint probability mass function is,

P (0, 0) = \frac{{}_{2}C_{2} \cdot {}_{2}C_{0} \cdot {}_{1}C_{0}}{{}_{5}C_{2}} = 0.1

P (1, 0) = \frac{{}_{2}C_{1} \cdot {}_{2}C_{1} \cdot {}_{1}C_{0}}{{}_{5}C_{2}} = 0.4

P (1, 1) = \frac{{}_{2}C_{0} \cdot {}_{2}C_{1} \cdot {}_{1}C_{1}}{{}_{5}C_{2}} = 0.2

P (2, 0) = \frac{{}_{2}C_{0} \cdot {}_{2}C_{2} \cdot {}_{1}C_{0}}{{}_{5}C_{2}} = 0.1

P (0, 1) = \frac{{}_{2}C_{1} \cdot {}_{2}C_{0} \cdot {}_{1}C_{1}}{{}_{5}C_{2}} = 0.2

Now, marginal probability function of X is,

P_{X} (X = 0) = 0.1 + 0.2 = 0.3

P_{X} (X = 1) = 0.4 + 0.2 = 0.6

P_{X} (X = 2) = 0.1 + 0 = 0.1

Now, marginal probability function of Y is,

P_{X} (Y = 0) = 0.1 + 0.4 + 0.1 = 0.6

P_{X} (Y = 1) = 0.2 + 0.2 + 0 = 0.4