# Which of the following events has the largest probability?

In a box there are 3 green, 3 yellow and 3 blue cards. For each colour cards are marked 1, 2 and 3. You take randomly three cards from the box. Which of the following events has the largest probability?
A) The three cards are of the same colour
B) The three cards, independently on their colours, have numbers 1, 2 and 3.
C) The three cards are of three different colours
D) The three cards have the same number.

I calculated the probability of each option as following:
A) 3/9 * 2/8 * 1/7 = 6/504
B) 3/9 * 3/8 * 3/7 = 27/504
C) 3/9 * 3/8 * 3/7 = 27/504
D) 3/9 * 2/8 * 1/7 = 6/504

Please clarify whether this is correct or not?

• It would he helpful to readers of your question if you explained how you obtained your answers. That makes it easier to detect any errors you may have made. Sep 22 at 11:13
• The answer given in the answer key is C. But here, according to your explanation, B is the correct answer Sep 22 at 11:47
• I obtained $1/28$ for both A and D and $9/28$ for both B and C. Sep 22 at 12:50
• What does "another answer" mean? Is that what you are supposed to choose when (as here) there are two events with the largest probability? It's not at all clear to me! Sep 22 at 13:47
• There is a symmetry between colours and numbers here. So A and D are equal, as are B and C. Sep 22 at 13:49

All answers are incorrect, I'll explain why:

A) multiply by $$3$$

B) multiply by $$3!$$

C) multiply by $$3!$$

D) multiply by $$3$$

The common mistake is that you consider only specific number/colour or specific ordered set of numbers/colours probability, while there are $$3$$ numbers and $$3!$$ possible ordered set of colours/numbers. So the correct reasoning for each answer is calculating the probability for a specific scenario (which is what you did) and then multiply by the number of possible scenarios, this is because the different scenarios represent events that are mutually exclusive (this means that if the $$3$$ cards are marked $$1,2,3$$ in order, they cannot be marked $$2,3,1$$ or any other combination). I'll give now a specific explenation of the mistakes:

Problems(A,D) and (B,C) are the same ($$(1,2,3)$$ or $$(y,g,b)$$ makes no difference)).

A,D) The probability of having $$card_1=x,card_2=x,card_3=x$$ where $$x$$ is a fixed number/colour is $$3/9*2/8*1/7$$, then you multiply by $$3$$ because $$x$$ could be $$1,2,3/y,g,b$$.

B,C) The probability of having for example $$card_1=1,card_2=2,card_3=3$$ is $$3/9*3/8*3/7$$, then you multiply by $$3!$$ because the cards can be $$(123,132,213,...)$$. Exact same thing with colours.

@SummertimeS4dness has explained your errors and how to correct them. Below, I have shown an alternative approach to the problems.

What is the probability that the three cards are of the same colour?

The probability that the first card is one of the three colours is $$1$$. The probability that the second card is of the same colour as the first is $$2/8$$. If those two cards are of the same colour, then the probability that the third card is of the same colour is $$1/7$$. Hence, the probability that the three cards are of the same colour is $$1 \cdot \frac{2}{8} \cdot \frac{1}{7} = \frac{1}{28}$$

What is the probability that the three cards, independently of colour, have numbers 1, 2, and 3?

The probability that the first card shows one of the numbers 1, 2, or 3 is $$1$$. The probability that the second card you select will have a different number than the number on the first card is $$6/8$$. The probability that the third card you select will have a different number than the first two cards given that those numbers are different is $$3/7$$. Hence, the probability that the three cards, independently of their colours, will have numbers 1, 2, and 3 is $$1 \cdot \frac{6}{8} \cdot \frac{3}{7} = \frac{9}{28}$$

What is the probability that the three cards are of different colours?

The probability that the first card is one of the three colours is $$1$$. The probability that the second card is of a different colour is $$6/8$$. If the first two cards are of different colours, the probability that the third card is a different colour from both of them is $$3/7$$. $$1 \cdot \frac{6}{8} \cdot \frac{3}{7} = \frac{9}{28}$$

What is the probability that the three cards have the same number?

The probability that the first card shows one of the three numbers 1, 2, or 3 is $$1$$. The probability that the second card will have the same number as the first is $$2/8$$. If the first two cards have the same number, then the probability that the third card has the same number as the first two is $$1/7$$. Hence, the probability that the three cards have the same number is $$1 \cdot \frac{2}{8} \cdot \frac{1}{7} = \frac{1}{28}$$

A) The chance of having all three cards of the same color is that the second card is the same color as the first one (whichever it was!) and the third one is the same color, too. After each draw there is one card less total in the pile and one card less of the color selected by the first card. So, it is $$P(A)=\frac {3-1}{9-1}\cdot\frac {3-2}{9-2} = \frac 2{8\cdot 7}=\frac 1{28}.$$

B) Here you want the second card to have a number different from the first one, and the third card to have a number different from them both. In the second draw six cards meet the requirement, and in the third one only three. So, $$P(B)=\frac 6{9-1}\cdot\frac 3{9-2}=\frac {18}{56}=\frac 9{28}.$$

C) This is equivalent to the B case, just use colors instead of numbers.

D) Now this is obvious: $$P(D)=\frac 28\cdot\frac 17=\frac 1{28}.$$

E) Does not make any sense to me. How could an "another answer" be "one of the following events", as the last sentence in the problem says?
BTW, the largest probability has, for example, the event you have three cards. This would make $$P(E)=1.$$ But that in no way relates to A) through D)...