4MA0 Higher January 2016 Paper 3 Q21
21 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.
(5)
Scheme 1 P(L,L) + P(L, NL) + P(NL, L)
Bus L = 0.07 ; BusNL = 0.63
BikeL = 0.015; BikeNL = 0.285
correct products may be seen instead of these values
| Scheme | Marks |
|---|---|
| 1 − 0.1 (=0.9) or 1 − 0.05 (=0.95) or 1 – 0.7 (=0.3) | M1 |
P(not late for work) = 0.7×0.9 + 0.3×0.95 (= 0.915 or \(\dfrac{183}{200}\) ) | M1 |
P(late for work) = 0.7×0.1 + 0.3×0.05 (= 0.085 or \(\dfrac{17}{200}\) ) OR 1 – “0.915” (=0.085) | M1 |
| P(late on at least one day) = 2 × "0.085"×"0.915" + "0.085"×"0.085" | M1 |
0.162775 or \(\dfrac{6511}{40000}\) oe | A1 |
| (5) |
Notes
M1: or at least 4 correct products from (addition not needed)
\(0.07^2 + 2 \times 0.07 \times 0.63 + 2 \times 0.07 \times 0.015 + 2 \times 0.07 \times 0.285 + 2 \times 0.63 \times 0.015 + 0.015^2 + 2 \times 0.015 \times 0.285\)
M1: or at least 8 correct products from (addition not needed)
\(0.07^2 + 2 \times 0.07 \times 0.63 + 2 \times 0.07 \times 0.015 + 2 \times 0.07 \times 0.285 + 2 \times 0.63 \times 0.015 + 0.015^2 + 2 \times 0.015 \times 0.285\)
M1: dep
or all 12 correct products with addition :
\(0.07^2 + 2 \times 0.07 \times 0.63 + 2 \times 0.07 \times 0.015 + 2 \times 0.07 \times 0.285 + 2 \times 0.63 \times 0.015 + 0.015^2 + 2 \times 0.015 \times 0.285\)
Scheme 2 1 – P(NL,NL)
BusNL = 0.63
BikeNL = 0.285
correct products may be seen instead of these values
| Scheme | Marks |
|---|---|
| 1 − 0.1 (=0.9) or 1 − 0.05 (=0.95) or 1 – 0.7 (=0.3) | M1 |
P(not late for work) = 0.7×0.9 + 0.3×0.95 (= 0.915 or \(\dfrac{183}{200}\) ) | M1 |
| P(not late both days) = "0.915" × "0.915" ( = 0.837225 ) | M1 |
| P(late on at least one day) = 1 − "0.837225"= 0.162775 | M1 |
0.162775 or \(\dfrac{6511}{40000}\) oe | A1 |
Notes
M1: or any two correct products from (addition not needed)
\(0.63^2 + 0.285^2 + 2 \times 0.63 \times 0.285\)
M1: dep
or all correct products (addition not needed)
\(0.63^2 + 0.285^2 + 2 \times 0.63 \times 0.285\)
(\(= 0.3969 + 0.081225 + 2 \times 0.17955 = 0.837225\))
M1: dep