Problem A

There are 52 cards in the deck, 13 of each suit - spades, hearts, diamonds and clubs. You pick 2 cards out of this deck.

What is the probability that you pick two spades?

Solution

"Classical" approach to this problem calls for calculating the total number of elementary events and the number of "good" elementary events (those that satisfy our condition).

The total number of events is the number of pairs out of 52 cards, that is C[52,2]=(52·51)/(1·2)=26·51.

The number of "good" events is the number of pairs out of 13 spades, that is C[13,2]=(13·12)/(1·2)=13·6

The probability, therefore, is (13·6)/(26·51)=1/17.

Problem B

There are 52 cards in the deck, 13 of each suit - spades, hearts, diamonds and clubs. You pick 1 cards out of this deck.

What is the probability that you pick a card higher in rank than 5 (that is, 6, 7, 8, 9, 10, Jack, Queen, King or Ace)?

Solution 1

This problem can be solved using the additive property of probability. The probability to pick a card of a rank 6 is 4/52=1/13 since there are four such cards in the deck (one for each suit). Exactly the same is the probability to pick a card of a rank 7 or 8, or 9, or 10 or Jack, or Queen, or King, or Ace.

These events are non-intersecting. Therefore, the probability of either one of them is a sum of their probabilities, that is

9·1/13=9/13.

Solution 2

This is, basically, the same solution. But, instead of calculating the probability of our event, we can calculate the probability of the complementary event and then subtract it from 1 (the probability of a certain or complete, or full event).

The complementary event is a combination of picking a card of a rank 2, 3, 4 or 5. There are 16 such cards in the deck (four each in each suit). Therefore, the probability of picking a card of a rank below 6 equals to 16/52=4/13.

The probability of our event will then be 1−4/13=9/13.

Problem C

We roll four dice.

What is the probability of rolling number 6 on, at least, one of them? Is it a good game for you if hitting number 6 at least once is a win? How about hitting 6 at least once on three dice?

Solution

It's easier to calculate the probability of not rolling number 6 on any of them. This is a combination of independent events of not rolling number 6 on each of four dice.

The probability of not rolling 6 on one dice is a sum of probabilities to roll 1 or 2, or 3, or 4, or 5, that is 5/6.

The combination of these events on four dice has a probability equal to a product of their probabilities, that is (5/6)^4.

The probability of a complementary event of rolling, at least, one number 6 is, therefore, 1−(5/6)^4≅0.5177.

This is greater than 1/2 and is a game you would more often win than lose.

With three dice the corresponding answer is 1−(5/6)^3≅0.4213, which is substantially less than 1/2 and definitely is a losing game for you.

Problem D

There are three manufacturers of T-shirts - A, B and C. All of them produce white and non-white T-shirts.

75% of T-shirts produced by A are white.

50% of T-shirts produced by B are white.

25% of T-shirts produced by C are white.

A department store sells T-shirt produced by all three manufacturers and none others.

25% of all T-shirts it purchases wholesale are from the manufacturer A.

35% of all T-shirts it purchases wholesale are from the manufacturer B.

40% of all T-shirts it purchases wholesale are from the manufacturer C.

If you randomly pick a T-shirt sold in this department store, what is the probability of it being white?

Hint

Use the formula of total probability:

P(X) = P(A∩X) + P(B∩X) + P(C∩X)

where events A, B and C are mutually exclusive and cover an entire sample space.

Solution

The event we are interested in can be represented as a sum of three events:

You pick a white T-shirt produced by the manufacturer A, you pick a white T-shirt produced by the manufacturer B or you pick a white T-shirt produced by the manufacturer C.

Let A be an event of picking a T-shirt produced by the manufacturer A,

B be an event of picking a T-shirt produced by the manufacturer B,

and C be an event of picking a T-shirt produced by the manufacturer C.

These events, A, B and C are mutually exclusive and their union covers an entire sample space.

Let X be an event of picking a white T-shirt - this is an event whose probability we have to find.

The representation we talk about can be expressed as the following formula (we use symbol ∩ to represent an intersection and symbol + to represent a union of events):

X=(A∩X)+(B∩X)+(C∩X)

Therefore, using a formula of total probability

P(X) = P(A∩X) + P(B∩X) + P(C∩X)

To find out each of these probabilities, we can use a formula of conditional probability: P(A∩X)=P(A)·P(X|A)

P(B∩X)=P(B)·P(X|B)

P(C∩X)=P(C)·P(X|C)

We know each one of these: P(A) = 0.25

P(B) = 0.35

P(C) = 0.40

P(X|A) = 0.75

P(X|B) = 0.50

P(X|C) = 0.25

This gives the result:

P(X) = 0.25·0.75 + 0.35·0.50 + 0.40·0.25 = 0.4625

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