Question 1124370: the probability that Henry is late for school is 0.3. find the probability on two consecutive he, never late. b) late only once
Answer by jim_thompson5910(35256)   (Show Source):
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If he's late 30% of the time, then he is not late 70% of the time
30% + 70% = 100%
0.3+0.7 = 1.0
Multiply 0.7 with itself: 0.7*0.7 = 0.49 = 49%
Therefore, the chances of him not being late 2 days in a row is 49%
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Define two events:
L = he's late
E = he's early or on time (a long-winded way of saying "he's not late")
P(L)+P(E) = 1 because there are only two options. Either event L happens or event E happens. The two events do not overlap.
Here is the sample space for two days
LL meaning he's late two days in a row
LE meaning he's late for day 1, but early or on time for day 2
EL meaning early or on time for day 1, but late for day 1
EE meaning he's early or on time for two days in a row
It's convenient to often write the sample space in curly braces like so:
{LL, LE, EL, EE}
The sample space is the set of all possible outcomes. For this problem, we're only restricting things to 2 days.
Earlier we computed P(EE) = 0.49 = 49%.
Note how:
P(E) = 1-P(L) = 1-0.3 = 0.7
and
P(EE) = P(E and E) = P(E)*P(E) = 0.7*0.7 = 0.49 = 49%
We want to find P(EL or LE), so we calculate the individual probabilities first
P(EL) = P(E)*P(L) = 0.7*0.3 = 0.21
P(LE) = P(L)*P(E) = 0.7*0.3 = 0.21
each event is independent, so multiplication is possible like this
Then add them up.
This addition is possible because EL and LE are mutually exclusive events.
P(EL or LE) = P(EL)+P(LE) = 0.21+0.21 = 0.42
He has a 42% chance of being late exactly once over the course of the two days.
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