Solve the given problem that involves Probability

In summary: Actually, I didn't omit...I just indicated the sum total from the first to last term... would yield same result. Apologies for confusion.
  • #1
chwala
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Homework Statement
See attached;
Relevant Equations
Probability
I may seek an alternative approach; actually i had thought that this would take a few minutes of my time..but just realized that it just takes a minute; My interest is only on highlighted part.

1677496127148.png


Text solution

1677496186076.png


My take;

##P(\text{at least one of the first three days is wet})=1-P(ddd)##
=## 1-(0.6×0.7×0.7)=0.706##

Of course the other way of doing it would also realize the same result but will need more time...i.e using
##P(www)+P(wwd)+P(wdw)+P(wdd)+P(dww)+...P(ddw)=0.706##

Cheers man!
 
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  • #2
chwala said:
My take;
##P(\text{at least one of the first three days is wet})=1-P(ddd)##
=## 1-(0.6×0.7×0.7)=0.706##
IMO, the approach above is the better approach. It uses the idea that the complement of "at least one day of the three is wet" is "none of the three days is wet." The probabilities of these two events has to add to 1.
chwala said:
Of course the other way of doing it would also realize the same result but will need more time...i.e using
##P(www)+P(wwd)+P(wdw)+P(wdd)+P(dww)+...P(ddw)=0.706##
The latter approach is trickier to get right in that you have to ensure that you have included all possible events in which one, two, or three of the days is wet. In the probabilities that you list, you have omitted one. These come from the following combinations.
3 wet days: ##\binom {3}{3} = 1## -- www
2 wet days: ##\binom{3}{2} = 3## -- wwd, wdw, dww
1 wet day: ##\binom{3}{1} - 3## -- wdd, dwd, ddw
 
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  • #3
Mark44 said:
IMO, the approach above is the better approach. It uses the idea that the complement of "at least one day of the three is wet" is "none of the three days is wet." The probabilities of these two events has to add to 1.
The latter approach is trickier to get right in that you have to ensure that you have included all possible events in which one, two, or three of the days is wet. In the probabilities that you list, you have omitted one. These come from the following combinations.
3 wet days: ##\binom {3}{3} = 1## -- www
2 wet days: ##\binom{3}{2} = 3## -- wwd, wdw, dww
1 wet day: ##\binom{3}{1} - 3## -- wdd, dwd, ddw
Actually, I didn't omit...I just indicated the sum total from the first to last term... would yield same result. Apologies for confusion.
 

FAQ: Solve the given problem that involves Probability

What is the probability of a single event occurring?

The probability of a single event occurring is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. Mathematically, it is expressed as P(A) = Number of favorable outcomes / Total number of possible outcomes.

How do you calculate the probability of multiple independent events occurring?

For multiple independent events, the probability of all events occurring is the product of their individual probabilities. If A and B are two independent events, then P(A and B) = P(A) * P(B).

What is the difference between independent and dependent events in probability?

Independent events are those whose outcomes do not affect each other. For example, flipping a coin and rolling a die. Dependent events are those where the outcome of one event affects the outcome of another. For example, drawing two cards from a deck without replacement.

How do you find the probability of either of two events occurring?

To find the probability of either of two events occurring, you use the formula: P(A or B) = P(A) + P(B) - P(A and B). This accounts for the overlap where both events might occur simultaneously.

What is a conditional probability and how is it calculated?

Conditional probability is the probability of an event occurring given that another event has already occurred. It is calculated using the formula: P(A|B) = P(A and B) / P(B), where P(A|B) is the probability of event A occurring given that event B has occurred.

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