The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
Suppose that you have 9 cards. 3 are green and 6 are yellow. The 3 green cards are numbered 1, 2, and 3. The 6 yellow cards are numbered 1, 2, 3, 4, 5 and 6. The cards are well shuffled. You randomly draw one card.
G = card drawn is green
Y = card drawn is yellow
E = card drawn is even-numbered
1) sample space
2) Enter probability P(G) as a fraction
3) P(G/E)
4) P(G and E)
5) P(G or E)
6) Are G and E are mutually exclusive ?
Given Information:
Number of green cards = 3
Number of yellow cards = 6
Total cards = 9
Required Information:
1) Sample Space = ?
2) P(G) = ?
3) P(G/E) = ?
4) P(G and E) = ?
5) P(G or E) = ?
6) Are G and E are mutually exclusive ?
Answer:
1) Sample Space = { G₁, G₂, G₃, Y₁, Y₂, Y₃, Y₄, Y₅, Y₆ }
2) P(G) = 1/3
3) P(G/E) = 1/4
4) P(G and E) = 1/9
5) P(G or E) = 2/3
6) Events G and E are not mutually exclusive.
Step-by-step explanation:
1)
The sample space is distinct number of all possible outcomes in a probability test.
When we randomly draw one card from the total 9 cards then every distinct possible outcome is given below
Sample Space = { G₁, G₂, G₃, Y₁, Y₂, Y₃, Y₄, Y₅, Y₆ }
2)
The probability of selecting green card is number of green cards divided by total number of cards,
P(G) = number of green cards/total cards
P(G) = 3/9
P(G) = 1/3
3)
The probability of selecting a green card given that the card is even numbered is the probability of number of green and even numbered cards divided by the probability of selecting a green card,
P(G/E) = P(G and E)/P(E)
How many cards are there which are green and also even numbered?
From the sample space we have G₂ is the only cards that is green and even numbered and the total cards are 9 so
P(G and E) = 1/9
We have 4 even numbered cards which are G₂, Y₂, Y₄, and Y₆ the total cards are 9 so
P(E) = 4/9
P(G/E) = P(G and E)/P(E)
P(G/E) = (1/9)/(9/4)
P(G/E) = 1/4
4)
The probability P(G and E) is already calculated and explained in previous part.
P(G and E) = 1/9
5)
The probability of selecting a card that is green or even numbered is the number of cards that are green or even numbered divided by total cards,
We have 3 cards that are green and 3 yellow cards which are even numbered so 3+3 = 6 cards and total cards are 9
P(G or E) = 6/9
P(G or E) = 2/3
6)
Mutually exclusive events:
When it is not possible for the two events to happen simultaneously then we say that they are mutually exclusive events.
For example:
If you toss a fair coin then is it possible that the heads and tails can appear simultaneously?
Yes you are right! they are mutually exclusive.
Now think about this, is it possible that a green card and even number card can be selected at the same time?
Yes you are right!
It is possible that the selected card is G₂ which is green and at the same time even number too.
So we can confidently say that events G and E are not mutually exclusive.
