Probability of a soccer game being played at night If I could get help with this problem, it would be greatly appreciated. I have been trying to use a truth table, but can't seem to solve it.
A soccer team plays $60\%$ of its games in the daytime and $40\%$ in the night. It wins $35\%$ of the night games and $50\%$ of the day games it plays. You hear that this team lost their last game. What is the probability that it was played at night?


*

*$0.752$

*$0.464$

*$0.125$

*$0.388$

 A: Hint:
Suppose the team plays $1000$ games


*

*How many does it expect to play in the day and how many at night?

*How many does it expect to lose in the day and how many at night?

*How many does it expect to lose in total?

A: What you're looking for in this problem is $P(night | loss)$. 
Bayes' theorem gives us the following:
$$P(night | loss) = \frac{P(night)*P(loss | night)}{P(loss)}$$
Two of these probabilities are given in the problem, the other is the denominator. You'll want to use the fact if B and C are mutually exclusive events that make up the whole sample space, meaning one or the other always occurs,
$$P(A) = P(A | B)*P(B) + P(A | C)*P(C)$$
This is known as the law of total probability.
A: Ok, never do that in an exam: The original probability of playing a game at night is $40\%$. But the information that the team lost the game increases the (conditional) probability that the game was played at night. Why? Because the team loses more games at night. So, exclude choices 3. and 4. that refer to a reduction of the initial probability.
And now the difficult part: How much does this information increase the probability? I expect not so much, so that I also exclude a posterior of $0.752$. So, if you had to guess ("do not try this at home") you should go with 2. $0.464$. 
A: 
A soccer team plays 60% of its games in the daytime and 40% in the night.

OK.

It wins 35% of the night games and 50% of the day games it plays.

Of all games 40% will be played at night and they will loose 65% of those.
All other losses will be the 60% played in daytime of which 50% will be lost.
Total percentage of all games lost will be ( ( 40% x 65% ) + ( 60% x 50% ) ) = 56%.
Total percentage of all games won bill be ( ( 40% x 35% ) + ( 60% x 50% ) ) = 44%
Which totals 100% ( as it should ).

You hear that this team lost their last game.
What is the probability that it was played at night?

Of the 56% of games lost in all, 40% x 65% will be lost at night.
Probability is $\frac{40 \% x 65 \%}{56 \%} \approx 46.429 \%$
Thanks to those that pointed our my ability to read does not match my ability to do maths and pointed our my mistake.  Apologies to all.
