Question

Consider randomly selecting a single individual and having that person test drive 3 different vehicles.

Answer 1

Answer:

(a) The probability that the individual likes both vehicle #1  and vehicle #2 is 0.40.

(b) The value of P (A₂ | A₃) is 0.7143.

(c) The events A₂ and A₃ are not independent.

(d) The probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.

Step-by-step explanation:

The events are defined as follows:

A₁ = an individual like vehicle #1

A₂ = an individual like vehicle #2

A₃ = an individual like vehicle #3

The information provided is:

$P(A_{1})=0.55\\P(A_{2})=0.65\\P(A_{3})=0.70\\P(A_{1}\cup A_{2})=0.80\\P(A_{2}\cap A_{3})=0.50\\P(A_{1}\cup A_{2}\cip A_{3})=0.88$

(a)

Compute the probability that the individual likes both vehicle #1  and vehicle #2 as follows:

$P(A_{1}\cap A_{2})=P(A_{1})+P(A_{2})-P(A_{1}\cup A_{2})\\=0.55+0.65-0.80\\=0.40$

Thus, the probability that the individual likes both vehicle #1  and vehicle #2 is 0.40.

(b)

Compute the value of P (A₂ | A₃) as follows:

$P(A_{2}|A_{3})=\frac{P(A_{2}\cap A_{3})}{P(A_{3}}\\=\frac{0.50}{0.70}\\=0.7143$

Thus, the value of P (A₂ | A₃) is 0.7143.

(c)

If two events X and Y are independent then,

$P(X\cap Y)=P(X)\times P(Y)\\P(X|Y)=P(X)$

The value of P (A₂ ∩  A₃) is 0.50.

The product of the probabilities, P (A₂) and P (A₃) is:

$P(A_{2})\times P(A_{3})=0.65\times0.70=0.455$

Thus, P (A₂ ∩ A₃) ≠ P (A₂) × P (A₃)

The value of P (A₂ | A₃) is 0.7143.

The value of P (A₂) is 0.65.

Thus, P (A₂ | A₃) ≠ P (A₂).

The events A₂ and A₃ are not independent.

(d)

Compute that probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ as follows:

$P(A_{2}\cup A_{3}|A_{1}^{c})=\frac{P((A_{2}\cup A_{3})\cap A_{1}^{c})}{P(A_{1}^{c})}\\=\frac{P((A_{2}\cup A_{3}\cup A_{1})-P(A_{1})}{1-P(A_{1})} \\=\frac{0.88-0.55}{1-0.55}\\=0.7333$

Thus, the probability that an individual likes at least one of A₂ and A₃ given they did not like A₁ is 0.7333.