# [SOLVED] Independent events 2

• Feb 9th 2009, 04:46 AM
nikk
[SOLVED] Independent events 2
Hai ...i confuse how to solve the problem as below.Is it the problem consider under independent event?

Question:
A die is thrown and the number from the set {1,2,3,4,5} is selected. Find the probabilities of

(i) B, the event in which the die shows an odd score,
ans : 1/2

my step:
B, the event in which the die shows an odd score
= { 1,3,5}
therefore P(B) = n(B)/ n(s)
= 3 / 5 : s = {1,2,3,4,5} , so n(s) = 5

(ii) C, the event in which the total score exceed 9,
ans : 1/10

(iii) D, the event in which the total score is 10
ans :1/15

hope some one can help me.

TQ
• Feb 9th 2009, 05:16 AM
Probability
Hello nikk
Quote:

Originally Posted by nikk
Hai ...i confuse how to solve the problem as below.Is it the problem consider under independent event?

Question:
A die is thrown and the number from the set {1,2,3,4,5} is selected. Find the probabilities of

(i) B, the event in which the die shows an odd score,
ans : 1/2

my step:
B, the event in which the die shows an odd score
= { 1,3,5}
therefore P(B) = n(B)/ n(s)
= 3 / 5 : s = {1,2,3,4,5} , so n(s) = 5

(ii) C, the event in which the total score exceed 9,
ans : 1/10

(iii) D, the event in which the total score is 10
ans :1/15

hope some one can help me.

TQ

If I understand the question correctly: in the first event, a die is thrown, resulting in a score from 1 to 6 being obtained; and then (independently) a number from 1 to 5 is chosen at random. Then:

(i) In your working n(s) = 6, not 5. We are simply looking at the throw of the die here. It has nothing to do with the second event. So $p(B) = \frac{3}{6} = \frac{1}{2}$

In the remaining parts of the question, I assume that the two numbers - one from the score on the die, and one from the set {1, 2, 3, 4, 5} are added together.

There are 30 possible outcomes now, when we combine the two events: each of the 6 possible scores on the die can be combined with each of the five numbers 1 through 5. If we represent these outcomes as ordered pairs (score on die, number from set), we have:

(ii) The total score exceeds 9 for the ordered pairs (5, 5), (6, 4) and (6, 5) only. Thus the probability is $\frac{3}{30} = \frac{1}{10}$.

(ii) The score is exactly 10 for the pairs (5, 5) and (6, 4). Thus the probability = $\frac{2}{30} = \frac{1}{15}$.

• Feb 9th 2009, 05:41 AM
nikk
tq
Quote:

Hello nikkIf I understand the question correctly: in the first event, a die is thrown, resulting in a score from 1 to 6 being obtained; and then (independently) a number from 1 to 5 is chosen at random. Then:

(i) In your working n(s) = 6, not 5. We are simply looking at the throw of the die here. It has nothing to do with the second event. So $p(B) = \frac{3}{6} = \frac{1}{2}$

In the remaining parts of the question, I assume that the two numbers - one from the score on the die, and one from the set {1, 2, 3, 4, 5} are added together.

There are 30 possible outcomes now, when we combine the two events: each of the 6 possible scores on the die can be combined with each of the five numbers 1 through 5. If we represent these outcomes as ordered pairs (score on die, number from set), we have:

(ii) The total score exceeds 9 for the ordered pairs (5, 5), (6, 4) and (6, 5) only. Thus the probability is $\frac{3}{30} = \frac{1}{10}$.

(ii) The score is exactly 10 for the pairs (5, 5) and (6, 4). Thus the probability = $\frac{2}{30} = \frac{1}{15}$.

tq for your explanation....base on your explanation...(There are 30 possible outcomes now, when we combine the two events: each of the 6 possible scores on the die can be combined with each of the five numbers 1 through 5. If we represent these outcomes as ordered pairs (score on die, number from set)

Question (i) B, the event in which the die shows an odd score,
ans : 1/2

so what i understand here is the outcomes should be (number from set which is odd, score on die which is odd)
B = { (1,1), (1,3) , (1, 5),
(2,1), (2,3) , (2, 5),
(3,1), (3,3) , (3, 5),
(4,1), (4,3) , (4, 5),
(5,1), (5,3) , (5, 5) }
so, n(B) = 15 n(s) = 30

so P(B) 15/30 = 1/2

is it wright ??
• Feb 9th 2009, 10:59 AM
Probability
Hello nikk
Quote:

Originally Posted by nikk
tq for your explanation....base on your explanation...(There are 30 possible outcomes now, when we combine the two events: each of the 6 possible scores on the die can be combined with each of the five numbers 1 through 5. If we represent these outcomes as ordered pairs (score on die, number from set)

Question (i) B, the event in which the die shows an odd score,
ans : 1/2

so what i understand here is the outcomes should be (number from set which is odd, score on die which is odd)
B = { (1,1), (1,3) , (1, 5),
(2,1), (2,3) , (2, 5),
(3,1), (3,3) , (3, 5),
(4,1), (4,3) , (4, 5),
(5,1), (5,3) , (5, 5) }
so, n(B) = 15 n(s) = 30

so P(B) 15/30 = 1/2

is it wright ??

Yes, you can explain it like this, where you list all the pairs of scores in which the die shows an odd number. As you've shown, this results in 15 favourable outcomes out of a total of 30, giving a probability of one-half.

But a much simpler explanation is to use your original method, where you just look at the score on the die, and say that there are 3 odd scores out of a total of 6, again giving a probability of one-half, because the number chosen from the set of 5 is not relevant in this part of the question.