Question

A service station has both self-service and full-service
islands. On each island, there is a single regular unleaded pump
with two hoses. Let *X* denote the number of hoses being
used on the self-service island at a particular time, and let
*Y* denote the number of hoses on the full-service island in
use at that time. The joint pmf of *X* and *Y*
appears in the accompanying tabulation.

y |
||||

,
x)y |
0 | 1 | 2 | |

x |
0 | 0.10 | 0.03 | 0.02 |

1 | 0.06 | 0.20 | 0.08 | |

2 | 0.06 | 0.14 | 0.31 |

(a) What is *P*(*X* = 1 and *Y* = 1)?

*P*(*X* = 1 and *Y* = 1) =

(b) Compute *P*(*X* ≤ 1 and *Y* ≤ 1).

*P*(*X* ≤ 1 and *Y* ≤ 1) =

(c) Give a word description of the event {*X* ≠ 0 and
*Y* ≠ 0}.

At most one hose is in use at both islands.At least one hose is in use at both islands. One hose is in use on both islands.One hose is in use on one island.

Compute the probability of this event.

*P*(*X* ≠ 0 and *Y* ≠ 0) =

(d) Compute the marginal pmf of *X*.

x |
0 | 1 | 2 |

p_{X}(x) |

Compute the marginal pmf of *Y*.

y |
0 | 1 | 2 |

p_{Y}(y) |

Using *p*_{X}(*x*), what is
*P*(*X* ≤ 1)?

*P*(*X* ≤ 1) =

(e) Are *X* and *Y* independent rv's? Explain.

*X* and *Y* are not independent because
*P*(*x*,*y*) =
*p*_{X}(*x*) ·
*p*_{Y}(*y*).*X* and
*Y* are independent because
*P*(*x*,*y*) =
*p*_{X}(*x*) ·
*p*_{Y}(*y*). *X*
and *Y* are independent because
*P*(*x*,*y*) ≠
*p*_{X}(*x*) ·
*p*_{Y}(*y*).*X* and
*Y* are not independent because
*P*(*x*,*y*) ≠
*p*_{X}(*x*) ·
*p*_{Y}(*y*).

Answer #1

Here is the table of probability is

p(x,y) | y=0 | y=1 | y=2 | Total |

x=0 | 0.1 | 0.03 | 0.02 | 0.15 |

x=1 | 0.06 | 0.2 | 0.08 | 0.34 |

x=2 | 0.06 | 0.14 | 0.31 | 0.51 |

Total | 0.22 | 0.37 | 0.41 | 1 |

*P*(*X* = 1 and *Y* = 1) = 0.20

(b) *P*(*X* ≤ 1 and *Y* ≤ 1) = P(X = 0, Y =
0) + P(X = 1, Y = 0) + P(X = 0, Y = 1) + P(X = 1, Y = 1)

= 0.1 + 0.03 + 0.06 + 0.2 = 0.39

(c) event {*X* ≠ 0 and *Y* ≠ 0} is At
least one hose is in use at both islands

*P*(*X* ≠ 0 and *Y* ≠ 0) = 1 - [P(X = 0) +
P(Y = 0) - P(X = 0, Y = 0)]

= 1 - [0.15 + 0.22 - 0.1] = 0.73

(d) Marginal pmf of X is

p(x) = 0.15 ; x= 0

= 0.34; x = 1

= 0.51; x = 2

Here

P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.15 + 0.34 = 0.49

Marginal pmf of Y is

p(y) = 0.22 ; Y = 0

= 0.37; Y = 1

= 0.41 ; Y = 2

(e) as we can see that

f(x) * f(y)
f(x,y) as *X* and *Y* are not independent because
*P*(*x*,*y*) ≠
*p**X*(*x*) ·
*p**Y*(*y*).

A service station has both self-service and full-service
islands. On each island, there is a single regular unleaded pump
with two hoses. Let X denote the number of hoses being used on the
self-service island at a particular time, and let Y denote the
number of hoses on the full-service island in use at that time. The
joint pmf of X and Y appears in the accompanying tabulation. y p(x,
y) 0 1 2 x 0 0.10 0.05 0.01 1...

A service station has both self-service and full-service
islands. On each island, there is a single regular unleaded pump
with two hoses. Let X denote the number of hoses being
used on the self-service island at a particular time, and let
Y denote the number of hoses on the full-service island in
use at that time. The joint pmf of X and Y
appears in the accompanying tabulation.
y
p(x,
y)
0
1
2
x
0
0.10
0.03
0.01 ...

A service station has both self-service and full-service
islands. On each island, there is a single regular unleaded pump
with two hoses. Let X denote the number of hoses being used on the
self-service island at a particular time, and let Y denote the
number of hoses on the full-service island in use at that time. The
joint pmf of X and Y appears in the accompanying tabulation.
P(X=x,Y=y) Y X 0 1 2 0 0.1 0.04 0.02 1 0.08...

A service station has both self-service and full-service
islands. On each island, there is a single regular unleaded pump
with two hoses. Let X denote the number of hoses being used on the
self-service island at a particular time, and let Y denote the
number of hoses on the full-service island in use at that time. The
joint pmf of X and Y appears in the accompanying tabulation. y p(x,
y) 0 1 2 x 0 0.10 0.03 0.01 1...

The joint probability distribution of the number X of
cars and the number Y of buses per signal cycle at a
proposed left-turn lane is displayed in the accompanying joint
probability table.
y
p(x,
y)
0
1
2
x
0
0.010
0.015
0.025
1
0.020
0.030
0.050
2
0.050
0.075
0.125
3
0.060
0.090
0.150
4
0.040
0.060
0.100
5
0.020
0.030
0.050
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problem 1, to compute
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