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Probability question mark scheme

So in a past paper, there's this question:
Peter travels to work either by bus or by bike. The probability that Peter will travel to work by bus on any one day is 0.7 Whenever Peter travels to work by bus, the probability that he will be late is 0.1 Whenever Peter travels to work by bike, the probability that he will be late is 0.05 Peter is going to go to work on Monday and on Tuesday. Work out the probability that he will be late for work on at least one of these days.

So I try solve it by doing ((0.7x0.1) + (0.3 x 0.05)), which would give the probability of being late. Then I do ((0.7x0.1) + (0.3 x 0.05)) x 2 to make up for the fact that there are two days. My answer is 0.17. However, the mark scheme (attached) doesn't seem to accept this. What's going on?
It's because he could get the bus on Monday and use his bike Tuesday or he could get the bus both days or use his bike both days.. And then you've got to work out the probability of him being late with the bike, and then being late with the bus, but not the other opinion, and also the probability of being late with both.
(edited 6 years ago)
If we use "L" to denote "Late" and "NL" to denote "Not Late", there are four things that can happen over the two days:

(L, L)
(L, NL)
(NL, L)
(NL, NL)

The sum of the probabilities of all of the above is 1.

The probability of being late on at least one of the two days is P(L, L) + P(L, NL) + P(NL, L). As P(L, NL) is the same as P(NL, L), this can be re-written as P(L, L) + 2P(L, NL). Or, as they've done in the mark scheme, the probability of being late on at least one of the two days is 1 - P(NL, NL).

So why does your approach give the wrong answer? You've correctly worked out the value of P(L), but you need to consider that when you look over the two days, P(L) = P(L, L) + P(L, NL). If in doubt about that, draw a new tree with L and NL as its branches, with two stages. From that it follows that 2P(L) = 2P(L,L) + 2P(L, NL). The upshot is that you've ended up double-counting the (L, L) case so your answer is a little on the high side.

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