Respuesta :
Answer:
Step-by-step explanation:
When two cubes are rolled then the favourable outcomes are: 36 (given below)
(1,1) ; (1,2) ; (1,3) ;(1,4) ; (1,5) ;(1,6)
(2,1) ; (2,2) ; (2,3) ;(2,4) ; (2,5) ;(2,6)
(3,1) ; (3,2) ; (3,3) ;(3,4) ; (3,5) ;(3,6)
(4,1) ; (4,2) ; (4,3) ;(4,4) ; (4,5) ;(4,6)
(5,1) ; (5,2) ; (5,3) ;(5,4) ; (5,5) ;(5,6)
(6,1) ; (6,2) ; (6,3) ;(6,4) ; (6,5) ;(6,6)
Favourable events of numbers on both cubes that give a sum less than 10 are (1,1) ; (1,2) ; (1,3) ;(1,4) ; (1,5) ;(1,6) ;(2,1) ; (2,2) ; (2,3) ;(2,4) ; (2,5) ;(2,6) ;(3,1) ; (3,2) ; (3,3) ;(3,4) ; (3,5) ;(3,6); (4,1) ; (4,2) ; (4,3) ;(4,4) ; (4,5);(5,1) ; (5,2) ; (5,3) ;(5,4) ;(6,1) ; (6,2) ; (6,3)
so total no. of favourable events of numbers on both cubes that give a sum less than 10 = 30
Probability of getting sum less than 10 = [tex]\frac{30}{36}[/tex]
Favourable events of numbers on both cubes that give a sum that is a multiple of 3 are (1,3);(1,6);(2,3);(2,6);(3,1) ; (3,2) ; (3,3) ;(3,4) ; (3,5) ;(3,6);(4,3);(4,6);(5,3);(5,6);(6,1) ; (6,2) ; (6,3) ;(6,4) ; (6,5) ;(6,6)
so total no. of favourable events of numbers on both cubes that give a sum that is a multiple of 3 = 20
Probability of getting a multiple of 3=[tex]\frac{20}{36}[/tex]
(A) P(B|A)= \frac{P(B∩A)}{P(A)} = [tex]\frac{15/36}{30/36}[/tex] = [tex]\frac{15}{30}[/tex]=0.5
(B)P(A|B)=[tex] \frac{P(A∩B)}{P(B)}[/tex]= [tex]\frac{15/36}{5/9}[/tex]=0.75
(C) {A∩B} = {3, 6, 9, 12, 15, 18}
(D) {A} = {1, 2, 3, 4, 5 ,6, 7, 8, 9}l
The conditional probability of B given that A occurs first is P( B / A ) = 15 /10.
Complete the conditional-probability formula for event B given that event A occurs first by writing A and B is P( B / A ) = {P(B∩A)} / {P(A)}.
The members of the sample space for event A ∩ B is {A∩B} = {3, 6, 9, 12, 15, 18}
The members of the sample space for event A {A} = {1, 2, 3, 4, 5 ,6, 7, 8, 9}l.
Conditional probability
Conditional probability is the probability of one event occurring with some relationship to one or more other events.
Given information
Two number cubes are rolled for two separate events:
All of the combinations of numbers on both cubes give a sum less than 10.
All combinations of numbers on both cubes give a sum that is a multiple of 3.
The statements are as follows;
1. In terms of a reduced fraction, find the conditional probability of B given that A occurs first is;
[tex]\rm P(\dfrac{B}{A}})=\dfrac{P(B\cap A)}{P(A)} =\dfrac{\dfrac{15}{36}}{\dfrac{10}{36}}\\\\\dfrac{P(B\cap A)}{P(A)} ={\dfrac{15}{36}}\times {\dfrac{36}{10}}\\\\\dfrac{P(B\cap A)}{P(A)} =\dfrac{15}{10}[/tex]
The conditional probability of B given that A occurs first is P( B / A ) = 15 /10.
2, Complete the conditional-probability formula for event B given that event A occurs first by writing A and B in the blanks:
P( B / A ) = {P(B∩A)} / {P(A)}
3. List the members of the sample space for event A ∩ B.
{A∩B} = {3, 6, 9, 12, 15, 18}
4. The members of the sample space for event A {A} = {1, 2, 3, 4, 5 ,6, 7, 8, 9}l.
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