## Wednesday, March 4, 2015

### Unizor - Probability - Easy Problems 5

Problem A

In the box there are R red balls and B black ones. If we randomly pick N balls from the box, what is the probability to have S red balls among them?

Solution

Our sample space contains all the combinations of N balls out of R+B, that is there are C[R+B,N] combinations of equal probability of occurrence.

Out of these elementary events we need only those combinations that contain exactly S red balls with the rest N−S balls being black. Since there are R red balls, we have C[R,S] choices for a group of S red balls. With each of them, similarly, we have C[B,N−S] choices for a group of N−S black balls. Therefore, the number of choices for a group of N balls that contains S red and N−S black balls equals to C[R,S]·C[B,N−S].

Hence, the probability to pick such a group is

C[R,S]·C[B,N−S] / C[R+B,N].

Problem B

Standard deck of cards contains 52 cards of four different suits. Let's call cards ranking from 2 to 10 Numerics and others (Jacks, Queens, Kings and Aces) we will call Pictures.

So, we have 36 Numerics and 16 Pictures in the deck.

We randomly pull 2 cards out of deck.

Calculate the probabilities of having

(a) both Numerics;

(b) both Pictures;

(c) one Numeric and one Picture.

Solution

Our sample space contains C[52,2] pairs of cards. All are equally probable.

(a) The number of pairs with both Numeric cards equals to C[36,2]. Therefore, the probability of pulling a pair of Numerics is C[36,2] / C[52,2] ≅ 0.4751.

(b) The number of pairs with both Picture cards equals to C[16,2]. Therefore, the probability of pulling a pair of Numerics is

C[16,2] / C[52,2] ≅ 0.0905.

(c) The number of pairs with one Numeric and one Picture cards equals to C[36,1]·C[16,1]. Therefore, the probability of pulling a pair of Numerics is C[36,1]·C[16,1] / C[52,2] ≅ 0.4344.

NOTE: It's always a good practice to check that sum of probabilities of all mutually exclusive events equals to 1.

In our case all three events are mutually exclusive and encompass all possible results of the experiments. Their probabilities share the same denominator C[52,2]. Let's check that the sum of all three numerators equals to this denominator.

C[52,2] = 52·51/(1·2) = 1326

C[36,2] = 36·35/(1·2) = 630

C[16,2] = 16·15/(1·2) = 120

C[36,1]·C[16,1] = 36·16 = 576

630+120+576 = 1326

Check.

Problem C

The revolver can hold 7 bullets, but there is only one bullet in it. A person playing "Russian roulette" spins the cylinder and pulls the trigger. Then he does the same once more.

What is the probability of not firing these two times in a row?

Solution

The probability of firing is 1/7. The probability of not firing is, therefore, 6/7.

Two events, not firing the first time and not firing the second time, are independent, and the probability of each is 6/7. The probability of a combination of these two independent events equals to a product of their probabilities, that is

(6/7)·(6/7)=36/49.

Problem D

Knights of the Round Table, including Sir Lancelot and King Arthur himself, randomly choose their chairs at the round table in the King Arthur's castle.

Assuming there are N knights, what is the probability of Lancelot and King Arthur to sit next to each other?

Solution 1

Let King Arthur choose whatever chair he wants. Then there are N−1 chairs left for Sir Lancelot. Only 2 of them are next to King Arthur. Therefore, the probability to sit next to him is 2/(N−1).

Solution 2

The total number of positions N knights can take around the table is N!. The number of positions Sir Lancelot and King Arthur can take to be next to each other is 2·N (any chair out of N for one and two positions of another on both sides of the first). With each of them there are (N−2)! positions of other N−2 knights. Therefore, the number of positions all knights can take with Sir Lancelot and King Arthur sitting next to each other equals to

2·N·(N−2)! / N! = 2/(N−1)

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