Juan goes to college each day by any one of car or bus or walking. The probability that he goes by car is $$\(0.2\)$$, the probability that he goes by bus is $$\(0.45\)$$ and the probability that he walks is $$\(0.35\)$$. When Juan goes by car, the probability that he arrives early is $$\(0.6\)$$. When he goes by bus, the probability that he arrives early is $$\(0.1\)$$. When he walks he always arrives early. Find the probability that Juan goes to college by car given that he arrives early. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mathematics
IGCSE&ALevel
CAIE
Exam No:9709_s20_qp_53 Year:2020 Question No:1(b)

Answer:

$
\mathrm{P}(\mathrm{C} \mid \mathrm{E})=\frac{\mathrm{P}(\mathrm{C} \cap \mathrm{E})}{\mathrm{P}(\mathrm{E})}=\frac{0.2 \times 0.6}{0.2 \times 0.6+0.45 \times 0.1+0.35 \times 1}
$

Summing three appropriate 2-factor probabilities
\(\frac{0.12}{0.515}\)
\(0.233\) or \(\frac{12}{515}\)

Knowledge points:

5.3.4 calculate and use conditional probabilities in simple cases.

Solution:

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