Answer :
Answer:
(a) [tex]0.099[/tex].
(b) [tex]0.48[/tex].
Step-by-step explanation:
Under the assumption that the three events are independent, the probability that all of them happen is equal to the product of the probability of the three individual events:
[tex]\begin{aligned}& P(\text{all three}) \\ =\; & P(\text{S $\cup$ M $\cup$ L}) \\ =\; & P(\text{S})\, P(\text{M})\, P(\text{L}) && (\text{S, M, and L are independent})\\ =\; & 0.66 \times 0.5 \times 0.3 \\=\; & 0.099\end{aligned}[/tex].
The probability that a event does not happen is equal to [tex]1[/tex] minus the probability that this event does happen. For example, given that the probability [tex]\text{L}[/tex] bought a DVD is [tex]0.3[/tex], the probability [tex]\text{L}[/tex] does not buy a DVD would be [tex](1 - 0.3)[/tex]:
[tex]P(\text{$\lnot$L}) = 1 - P(\text{L}) = 1 - 0.3[/tex].
To find the probability that at least two of the events happen, observe that this event can be decomposed into the following events that are mutually exclusive:
- Exactly two of the three bought a DVD.
- All three bought a DVD.
These two events are mutually exclusive because if one of them happens, the other event in the list would not happen. Moreover, the probability that one of the mutually-exclusive events happen is equal to the sum of the probability of the individual events. This property makes it possible to find the probability of the event of interest (at least two bought a DVD) by decomposing that event into some mutually-exclusive sub-events (exactly two bought a DVD, or exactly three bought a DVD) where probabilities are easier to find:
[tex]\begin{aligned}& P({\rm \text{at least two}}) \\ =\; & P((\text{exactly two})\cup ({\rm \text{exactly three}})) \\ =\; & P(\text{exactly two}) + P(\text{exactly three}) && (\text{mutually exclusive})\end{aligned}[/tex]
Likewise, the probability that exactly two bought a DVD can be further decomposed:
- [tex]\text{S}[/tex] and [tex]\text{M}[/tex], but not [tex]\text{L}[/tex]: [tex]0.66 \times 0.5 \times (1 - 0.3)[/tex].
- [tex]\text{S}[/tex] and [tex]\text{L}[/tex], but not [tex]\text{M}[/tex]: [tex]0.66 \times 0.5 \times (1 - 0.3)[/tex].
- [tex]\text{M}[/tex] and [tex]\text{L}[/tex], but not [tex]\text{S}[/tex]: [tex]0.66 \times 0.5 \times (1 - 0.3)[/tex].
In other words:
[tex]\begin{aligned} & P(\text{exactly two}) \\=\; & P(\text{(S $\cap$ M $\cap$ ($\lnot$L)) $\cup$ (S $\cap$ ($\lnot$M) $\cap$ L) $\cup$ (($\lnot$S) $\cap$ M $\cap$ L})) \\ =\; &P(\text{S $\cap$ M $\cap$ ($\lnot$L))} \\ &+ P(\text{S $\cap$ ($\lnot$M) $\cap$ L}) \\ &+ P(\text{($\lnot$S) $\cap$ M $\cap$ L}}) \\ =\; & 0.66 \times 0.5 \times (1 - 0.3) \\ & + 0.66 \times (1 - 0.5) \times 0.3 \\ & + (1 - 0.66) \times 0.5 \times 0.3 \\ =\; & 0.381\end{aligned}[/tex].
Since it is already deduced that [tex]P(\text{all three}) = 0.099[/tex], the probability that at least two of the three buy a DVD would be:
[tex]\begin{aligned}& P({\rm \text{at least two}}) \\ =\; & P(\text{exactly two}) + P(\text{exactly three}) \\ =\; & 0.381 + 0.099 \\ =\; & 0.48\end{aligned}[/tex].
