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.

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 ?

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.

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step-by-step explanation:

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