I am trying to find the proportion of nhl 2k (video game) players that play more than 3 games or play on playstation. 40% of players use xbox, 50% use playstation, and 10% use pc. Of those who play on xbox, 60% play more than 3 games. Of those who play on playstation, 15% play more than 3 games. Of those who play on pc, 0% play more than 3 games.
What proportion of players play more than 3 games OR play on playstation?
I'm not sure if I am supposed to add these 2 proportions together and if my math is correct:
Probability of players play on XBOX = 0.4
Probability of players play on Playstation = 0.5
Probability of players play on pc = 0.1
Probability of players play more than 3 games on XBOX = 0.6
Probability of players play more than 3 games on Playstation = 0.15
Probability of players play more than 3 games on pc = 0.0
Probability of players play more than 3 games = 0.4 * 0.6 + 0.5 * 0.15 = 0.315
So 0.315 proportion of players play more than 3 games.
proportion of players who play on playstation = 0.5 * 0.315 = 0.1575
Should I be adding the probability of players that play more than 3 games to the proportion of players that play on playstation?
Firstly, some semantical notes. You state 'Probability of players play more than 3 games on XBOX = 0.6' which I am taking to mean 'Probability that a player plays more than three games on XBOX = 0.6' or 'Proportion of players that play more than three games on XBOX = 0.6'. Both of these are not correct. It would be correct to say 'Proportion of XBOX players that play more than three games = 0.6' or 'Probability that a player plays more than three games GIVEN that they play on XBOX = 0.6'. The key difference is the conditionality. The problem is saying that OF THE 40% that play XBOX, 60% play more than three games. This means that 'Probability that a player plays more than three games on XBOX' or 'Proportion of players that play more than three games on XBOX' is 0.4$\cdot$0.6 = 0.24. The same goes for your statements about Playstation and Other. You seem to understand this even if your statements are incorrect because you correctly calculate 'Probability that a player plays more than three games' to be 0.4$\cdot$0.6 + 0.5$\cdot$0.15 + 0.1$\cdot$0 = 0.315. I am confused by your last calculation. I don't know of any reason you would multiply those two figures together. You state that the problem asks for the probability that a player plays Playstation OR more than three games. To calculate this you should use the following formula: $$P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)$$ Here we are looking to calculate $P(\text{plays Playstation or plays more than three games})$ so our formula tells us this is $$P(\text{plays Playstation})+P(\text{plays more than three games})-P(\text{plays Playstation and plays more than three games})$$ We are given that, $$P(\text{plays Playstation})=0.5$$ And we computed, $$P(\text{plays Playstation and plays more than three games})=0.5\cdot0.15=0.075$$ $$P(\text{plays more than three games})=0.315$$ Therefore $$P(\text{plays Playstation or plays more than three games})=0.5+0.315-0.075=0.74$$ which is our answer.