Question

Suppose that you have 9 cards. 5 are green and 4 are yellow. The 5 green cards are numbered 1, 2, 3, 4, and 5. The 4 yellow cards are numbered 1, 2, 3, and 4. 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

Answer 1

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Suppose that you have 9 cards. 5 are green and 4 are yellow. The 5 green cards are numbered 1, 2, 3, 4, and 5. The 4 yellow cards are numbered 1, 2, 3, and 4. 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 = 5

Number of yellow cards = 4

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, G₄, G₅, Y₁, Y₂, Y₃, Y₄ }

2) P(G) = 5/9

3) P(G/E) = 1/2

4) P(G and E) = 2/9

5) P(G or E) = 7/9

6) Events G and E are not mutually exclusive.

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, G₄, G₅, Y₁, Y₂, Y₃, Y₄ }

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) = 5/9

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 and G₄ which are the only cards which are green and even numbered and the total cards are 9 so

P(G and E) = 2/9

We have 4 even numbered cards which are G₂, G₄, Y₂, Y₄ and the total cards are 9 so

P(E) = 4/9

P(G/E) = P(G and E)/P(E)

P(G/E) = (2/9)/(9/4)

P(G/E) = 2/4

P(G/E) = 1/2

4)

The probability P(G and E) is already calculated and explained in previous part.

P(G and E) = 2/9

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 5 green cards and 2 yellow cards which are even numbered so 5+2 = 7 cards and total cards are 9

P(G or E) = 7/9

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 or G₄ which are green and at the same time even number too.

So we can confidently say that events G and E are not mutually exclusive.

Answer 2

1. S = {G1, G2, G3, G4, G5, Y1, Y2, Y3, Y4}

2.
`P(G)=(5)/(9)`

3.
`P(G/E)=(1)/(2)`

4. P(G∩E)
`=(2)/(9)`

5. P(G∪E)
`=(7)/(9)`

Explanation:

1. The sample space are all the posibles option that we have when we draw one card, so it is equal to get:

S = {G1, G2, G3, G4, G5, Y1, Y2, Y3, Y4}

2. the probability of G, that the card drawn is green is equal to:

`P(G)=(5)/(9)`

Because, we have 9 cards and 5 of them are green.

3. The probability P(G/E) that the card is green given that the card drawn is even-numbered is calculated as:

P(G/E)=P(G∩E)/P(E)

So, the probability of get a even-numbered card is equal to:

`P(E)=(4)/(9)`

Because we have 9 cards and 4 of them are even-numbered. At the same way the probability P(G∩E) that a card is even and green is:

P(G∩E)
`=(2)/(9)`

Therefore, is equal to:

`P(G/E)=(2/9)/(4/9)=(1)/(2)`

4. The probability P(G∩E) the the card is green and it is a even-numbered is equal to
`(2)/(9)`

5. The probability P(G∪E) the the card is green or it is a even-numbered is calculated as:

P(G∪E)
`=(7)/(9)`

because we have 9 cards and 7 of them are green or are a even-numbered (G1, G2, G3, G4, G5, Y2, Y4)

6. Two events are mutually exclusive if one of them happens the other shouldn't happen. So, G and E aren't mutually exclusive because they both can happen at the same time, for example, if the card drawn is G4.