Day 2 - Probability, Compound Event Probability

Basic probability with dices

Problem

In this challenge, we practice calculating probability. In a single toss of 2 fair (evenly-weighted) six-sided dice, find the probability that their sum will be at most 9.

Mathematical explanation

A nice way to think about sums-of-two-dice problems is to lay out the sums in a 6-by-6 grid in the obvious manner.

We see that the identic values are on the same diagonal. The number of elements on the diagonal varies from 1 to 6 and then back to 1.

let's call A < x the event : the sum all the 2 tosses is at most x.

The value of \(P(A = i) = \frac{i-1}{36}\) if \(i \leq 7\) and \(P(A = i) = \frac{13-i}{36}\)

hence

Let's program it

sum([1 for d1 in range(1,7) for d2 in range(1,7) if d1+d2<=9]) / 36

0.8333333333333334

More dices

Problem

In a single toss of 2 fair (evenly-weighted) six-sided dice, find the probability that the values rolled by each die will be different and the two dice have a sum of 6.

Mathematical explanation

Let's consider 2 events : A and B. A compound event is a combination of 2 or more simple events. If A and B are simple events, then A∪B denotes the occurence of either A or B. A∩B denotes the occurence of A and B together.

We denote A the event "the values of each dice is different". The opposit event is A' "the values of each dice is the same".

We denote B the event "the two dice have a sum of 6", this probability has been computed on the first part of the article :

The probability of having 2 dice different of sum 6 is :

The probability that both A and B occure is equal to P(A∩B).

Since \(P(A|B)=\frac{P(A∩B)}{P(B)}\)

Let's program it

sum([1 for d1 in range(1,7) for d2 in range(1,7) if (d1+d2==6) and (d1!=d2)]) / 36

0.1111111111111111

Compound Event Probability

Problem

There are 3 urns labeled X, Y, and Z.

• Urn X contains 4 red balls and 3 black balls.
• Urn Y contains 5 red balls and 4 black balls.
• Urn Z contains 4 red balls and 4 black balls.

One ball is drawn from each of the urns. What is the probability that, of the 3 balls drawn, are 2 red and is 1 black?

Mathematical explanation

Let's write the different probabilities:

Addition rule

A and B are said to be mutually exclusive or disjoint if they have no events in common (i.e., and A∩B=∅ and P(A∩B)=0. The probability of any of 2 or more events occurring is the union (∪) of events. Because disjoint probabilities have no common events, the probability of the union of disjoint events is the sum of the events' individual probabilities. A and B are said to be collectively exhaustive if their union covers all events in the sample space (i.e., A∪B=S and P(A∪B)=1). This brings us to our next fundamental rule of probability: if 2 events, A and B, are disjoint, then the probability of either event is the sum of the probabilities of the 2 events (i.e., P(A or B) = P(A)+P(B))

Mutliplication rule

If the outcome of the first event (A) has no impact on the second event (B), then they are considered to be independent (e.g., tossing a fair coin). This brings us to the next fundamental rule of probability: the multiplication rule. It states that if two events, A and B, are independent, then the probability of both events is the product of the probabilities for each event (i.e., P(A and B)= P(A)xP(B)). The chance of all events occurring in a sequence of events is called the intersection (∩) of those events.

The balls drawn from the urns are independant hence :

p = P(2 red (R) and 1 back (B))

Each of those 3 probability if equal to the product of the probability of drawing each ball \(P(RRB) = P(R|X) * P(R|Y) * P(B|Z) = 4/7*5/9*1/2\)

• \(P(RRB) = 20/126\)

• \(P(RBR) = 16/126\)

• \(P(BRR) = 15/126\)

this leads to

• \(p = 51/126\)

and finally

Let's program it

X = 3*["B"]+4*["R"]
Y = 4*["B"]+5*["R"]
Z = 4*["B"]+4*["R"]
target = ["BRR", "RRB", "RBR"]

sum([1 for x in X for y in Y for z in Z if x+y+z in target])/sum([1 for x in X for y in Y for z in Z])

0.40476190476190477