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.

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.