Question

research has shown that 70% of a new small-medium enterprise is started by graduates while 30% are started by no graduates. it is also known that 60% of SMEs started by graduates are successful i.e they survived beyond 3 years, while 20% of those started by non-graduates are successful
(a)what is the probability that a new SMEs will not be successful/ (2)
(b)if it is known that a new SME is successful, what is the probability that it was started by a graduate? (2)

Answer 1

Let A be the event that a new small-medium enterprise is started by graduates.

Let B be the event that a new small-medium enterprise(SME) is started by no graduates.

Let E be the event that a new SME is successful.

Now, P(A) will be the probability that small-medium enterprise is started by graduates and

P(B) will be the probability that small-medium enterprise is started by no graduates.

P(E) will be the probability that a new SME is successful.

According to the question,

P(A) = 70%

P(B) = 30%

P(E/A) is the probability that a new SMU is successful given that it was started by graduates.

P(E/B) is the probability that a new SMU is successful given that it was started by no graduates.

$P(E \cap A)$ will be the probability that a new SMU started by graduates and is successful.

$P(E \cap A) = P(A) \times P(E/A)$

$P(E \cap A) = \dfrac{70}{100} \times \dfrac{60}{100}\\\Rightarrow P(E \cap A) = 42 \%$

Similarly,

$P(E \cap B) = \dfrac{30}{100} \times \dfrac{20}{100}\\\Rightarrow P(E \cap B) = 6 \%$

P(E) = 42% + 6%

P(E) = 48%

(a) Probability that a new SMEs will not be successful

$P(\overline{E}) = 1 - P(E)\\\Rightarrow P(\overline{E}) = 52\%$

(b) if it is known that a new SME is successful, the probability that it was started by a graduate:

$P(A/E) = \dfrac{P(A \cap E)}{P(E)}\\\Rightarrow \dfrac{42\%}{48\%}\\\Rightarrow 87.5\%$