96 Maˆ ıtriser l’al´ eatoire
60
40
20
15
30
45
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X M
X L
Fig. 4.2 – Domaine d’int´ egration de l’exercice 4.7.
4.7
Soient X M et X L les variables al´ eatoires donnant l’heure d’arriv´ ee (en minutes apr` es 20 h 00) de Marine et Laure. Elles suivent des lois uniformes respectivement de param` etres (15, 45) et (0, 60).
1. On cherche la probabilit´ e que les 2 heures d’arriv´ ee aient un ´ ecart inf´ erieur
` a 5 minutes, soit
P (| X L − X M |< 5) =
= P (−5 < X L − X M < 5) = P (X M − 5 < X L < X M + 5)
=
D
1
60
1
30
dx M dx L ,
o` u le domaine de d´ efinition D correspond `
a l’aire hachur´ ee repr´ esent´ ee
sur la figure 4.2. Ainsi
P (| X L − X M |< 5) =
45
15
xM +5
xM −5
1
30
1
60
dx L dx M
=
45
15
1
1800
(x M + 5 − x M + 5)dx M
=
1
180
45
15
dx M =
1
6
.
60
40
20
15
30
45
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X M
X L
Fig. 4.2 – Domaine d’int´ egration de l’exercice 4.7.
4.7
Soient X M et X L les variables al´ eatoires donnant l’heure d’arriv´ ee (en minutes apr` es 20 h 00) de Marine et Laure. Elles suivent des lois uniformes respectivement de param` etres (15, 45) et (0, 60).
1. On cherche la probabilit´ e que les 2 heures d’arriv´ ee aient un ´ ecart inf´ erieur
` a 5 minutes, soit
P (| X L − X M |< 5) =
= P (−5 < X L − X M < 5) = P (X M − 5 < X L < X M + 5)
=
D
1
60
1
30
dx M dx L ,
o` u le domaine de d´ efinition D correspond `
a l’aire hachur´ ee repr´ esent´ ee
sur la figure 4.2. Ainsi
P (| X L − X M |< 5) =
45
15
xM +5
xM −5
1
30
1
60
dx L dx M
=
45
15
1
1800
(x M + 5 − x M + 5)dx M
=
1
180
45
15
dx M =
1
6
.