The purpose of this series is developing understanding of stastical problems in AS and A level maths. Let's get started with a simple concept in statistics: Mutually exclusive events.

The probability of an OR event is calculated by the following rule:

`P(A || B) = P(A) + P(B) - P(A && B)`

The probability of event A or event B happening is equal to the probability of event A plus the probability of event B minus the probability of event A and event B happening simultaneously.

Mutually exclusive events are events that cannot happen at the same time. For example, the head and tail results of a toin coss are mutually exclusive because they can't both happen at once. Thus, the above example for a coin toss would look like this:

`P(H || T) = P(H) + P(T) - P(H && T)`

Note that the probaility of tossing a coin and the result being both head and tails is 0.

`P(H || T) = (0.5) + (0.5) - (0)`
`P(H || T) = 1`

Thus the probability of a coin toss result being a heads or a tails is 1, in other words: certain.


Your task:

You are going to have to work out the probability of one roll of a die returning two given outcomes, or rolls. Given that dice rolls are mutually exclusive, you will have to implement the above forumala. To make this interesting (this is a coding challenge after all), these dice are not fair and thus the probabilites of receiving each roll is different.

You will be given a two-dimensional array containing the number each of the results (1-6) of the die and the probability of that roll for example `[1 , 0.23]` as well as the two rolls for example `1` and `5`. 

Given the two roll probabilities to calculate, return the probability of a single roll of the die returning either. If the total probability of the six rolls doesn't add up to one, there is a problem with the die; in this case, return null. 
Return your result as a string to two decimal places.

Example below:

`1 : 1/6`

`2 : 1/6`

`3 : 1/6`

`4 : 1/6`

`5 : 1/6`

`6 : 1/6`

If asked for the rolls `1` and `2` then you would need to sum the probabilities, both `1/6` therefore `2/6` and return this. As above, you will need to return it as a decimal and not a  fraction.

Difficulty: EASY
Tags: Mathematics

Probability

Starter code:
def mutually_exclusive(dice, call1, call2):
	

Solutions:
def mutually_exclusive(dice, call1, call2):
	dice = dict(dice)
	if abs(sum(dice.values()) - 1) < 1e-12:
		return '%0.2f' % (dice[call1] + dice[call2])


def mutually_exclusive(dice, call1, call2):
	if sum((prob for (_, prob) in dice)) != 1:
		return None
	x = dict(dice)
	return format(x[call1] + x[call2], '.2f')


def mutually_exclusive(dice, call1, call2):
	d = dict(dice)
	return '{:0.2f}'.format(d[call1] + d[call2]) if 0.999 < sum(d.values()) < 1.001 else None


def mutually_exclusive(dice, call1, call2):
	prob = 0
	tprob = 0
	for i in dice:
		if i[0] == call1 or i[0] == call2:
			prob += i[1]
		tprob += i[1]
	return '{:.2f}'.format(prob) if tprob == 1 else None


def mutually_exclusive(dice, call1, call2):
	dice.sort()
	if sum((pair[1] for pair in dice)) != 1:
		return None
	else:
		return '{:.2f}'.format(dice[call1 - 1][1] + dice[call2 - 1][1])


def mutually_exclusive(dice, call1, call2):
	dice = {a: b for (a, b) in dice}
	return None if sum(dice.values()) != 1 else f'{float(dice[call1] + dice[call2]):.02f}'


def mutually_exclusive(dice, call1, call2):
	prob_of_each_roll = []
	for roll in dice:
		prob_of_each_roll.append(roll[1])
		if roll[0] == call1:
			p1 = roll[1]
		if roll[0] == call2:
			p2 = roll[1]
	if sum(prob_of_each_roll) != 1:
		return
	else:
		total_prob = p1 + p2
		return '{:.2f}'.format(total_prob)


def mutually_exclusive(dice, call1, call2):
	checknull = 0
	diecall1 = 0
	diecall2 = 0
	for die in dice:
		checknull += die[1]
		if die[0] == call1:
			diecall1 = die[1]
		elif die[0] == call2:
			diecall2 = die[1]
	if checknull != 1:
		return None
	else:
		return format(diecall1 + diecall2, '.2f')