In the previous Kata we discussed the OR case.

We will now discuss the AND case, where rather than calculating the probablility for either of two (or more) possible results, we will calculate the probability of receiving all of the viewed outcomes.

For example, if we want to know the probability of receiving head OR tails in two tosses of a coin, as in the last Kata we add the two probabilities together. However if we want to know the probability of receiving head AND tails, in that order, we have to work differently.

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

`P(A ∩ B) = P(A | B) * P(B)`

or

`P(B ∩ A) = P(B | A) * P(A)`

That is, the probability of A and B both occuring is equal to the probability of A given B occuring times the probability of B occuring or vice versa. 
If the events are mutually exclusive like in the case of tossing a coin, the probability of A occuring if B has occured is equal to the probability of A occuring by itself. In this case, the probability can be written as the below:

`P(A ∩ B) = P(A) * P(B)`

or

`P(B ∩ A) = P(B) * P(A)`

Applying to the heads and tails case:

`P(H ∩ T) = P(0.5) * P(0.5)`

or

`P(H ∩ T) = P(0.5) * P(0.5)`

The task:

You are given a random bag of 10 balls containing 4 colours. `Red`, `Green`, `Yellow` and `Blue`. You will also be given a sequence of 2 balls of any colour e.g. `Green` and `Red` or `Yellow` and `Yellow`. 

You have to return the probability of pulling a ball at random out of the bag and it matching the first colour and then pulling a ball at random out of the bag and it matching the second colour.

You will be given a boolean value of `true` or `false` which indicates whether the balls that is taken out in the first draw is replaced in to the bag for the second draw. Hint: this will determine whether the events are mutually exclusive or not.

You will receive two arrays and a boolean value. The first array contains the colours of the balls in the bag and the second contains the colour of the two balls you have to receive. As per above the final boolean value will indicate whether the bals arel being replaced `true` or not `false`.

Return the probability to 3 decimal places.

e.g. `[["y","r","g","b","b","y","r","g","r","r"],["r","b"],false]`


Other Kata in this series:
Statistics in Kata 1: OR case - Unfair dice

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Data structures

Mathematics

Probability

Starter code: def ball_probability(balls):
	

Solutions:
from collections import Counter as C

def ball_probability(b):
	(d, n, p) = (C(b[0]), len(b[0]), 1)
	for i in b[1]:
		p *= d.get(i, 0) / n
		n -= b[2] ^ 1
		d[i] -= b[2] ^ 1
	return round(p, 3)


def ball_probability(balls):
	(colors, (first, second), replaced) = balls
	first_choice = colors.count(first) / len(colors)
	second_choice = colors.count(second) / len(colors) if replaced else (colors.count(second) - 1 * (first == second)) / (len(colors) - 1)
	return round(first_choice * second_choice, 3)


def ball_probability(balls):
	size = len(balls[0])
	if not balls[2]:
		p1 = balls[0].count(balls[1][0])
		p2 = p1 - 1 if balls[1][0] == balls[1][1] else balls[0].count(balls[1][1])
		return round(p1 * p2 / (size * (size - 1)), 3)
	return round(balls[0].count(balls[1][0]) * balls[0].count(balls[1][1]) / size ** 2, 3)


def ball_probability(balls):
	(l, selection, replaced) = balls
	(a, b) = selection
	prob = l.count(a) * (l.count(b) + (0 if replaced or a != b else -1)) / (len(l) * (len(l) + (0 if replaced else -1)))
	return round(prob, 3)


def ball_probability(balls):
	(bag, draws, replaced) = balls
	return round(bag.count(draws[0]) / len(bag) * (bag.count(draws[1]) - (not replaced) * (draws[0] == draws[1])) / (len(bag) - (not replaced)), 3)


def ball_probability(balls):
	(bag, t, r) = balls
	(a, b) = (bag.count(i) for i in t)
	m = len(bag)
	return round(1.0 * a * (b if r or t[0] != t[1] else b - 1) / m / (m if r else m - 1), 3)


from collections import Counter

def ball_probability(balls):
	c = Counter(balls[0])
	(b1, b2) = balls[1]
	if balls[2]:
		return round(c[b1] / len(balls[0]) * c[b2] / len(balls[0]), 3)
	else:
		return round(c[b1] / len(balls[0]) * (c[b2] - 1 if b1 == b2 else c[b2]) / (len(balls[0]) - 1), 3)


def ball_probability(balls):
	(balls, draw, repl) = balls
	(b1, l1) = (balls.count(draw[0]), len(balls))
	if b1 == 0:
		return 0
	b2 = balls.count(draw[1]) if draw[0] != draw[1] else b1 + repl - 1
	l2 = l1 + repl - 1
	return round(b1 / l1 * b2 / l2, 3)


from collections import Counter

def ball_probability(balls):
	(bag, (call1, call2), replaced) = balls
	(bag, total) = (Counter(bag), len(bag))
	A = bag[call1] / total
	if not replaced:
		bag[call1] -= 1
		total -= 1
	B = bag[call2] / total
	return round(A * B, 3)