probability

Malachi returns at the end of the first day on which it rains.
For him to return at the end of monday, monday should be the frst day when it rains

Probability that it doesn't rain on Saturday = 1-0.2 = 0.8
Probability that it doesn't rain on Sunday = 1-0.2 = 0.8
Probability that it rains on Monday = 0.2
Probability that it rains on monday and not on sat and sun = 0.8*0.8*0.2 = 0.128

Option B

smagish wrote:On Saturday morning, Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains. If on the first three days of the vacation the probability of rain on each day is 0.2, what is the probability that Malachi will return
home at the end of the day on the following Monday?
A. 0.008
B. 0.128
C. 0.488
D. 0.512
E. 0.640

This question is the way-way difficult to get an answer rather reading from the proposed solution by riju The question can be solved with 100% guarantee of correctness if we use binomial probability distribution for a series of 'success'-'failure' events given P(failure=rain) is always the same 0.2

Thus, 3C1*(0.2)*(0.8)^2=0.384. This will be the sum of all possible events: Rain-Not Rain-Not Rain, Not Rain-Rain-Not Rain, Not Rain-Not Rain-Rain.

If you look into the condition I underscored in quote you should notice that the first two options in a series of events are locked (switched off). Hence we are left only with 0.128 which is 0.384/3 the third option.

pemdas wrote:

smagish wrote:On Saturday morning, Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains. If on the first three days of the vacation the probability of rain on each day is 0.2, what is the probability that Malachi will return
home at the end of the day on the following Monday?
A. 0.008
B. 0.128
C. 0.488
D. 0.512
E. 0.640

This question is the way-way difficult to get an answer rather reading from the proposed solution by riju The question can be solved with 100% guarantee of correctness if we use binomial probability distribution for a series of 'success'-'failure' events given P(failure=rain) is always the same 0.2

Thus, 3C1*(0.2)*(0.8)^2=0.384. This will be the sum of all possible events: Rain-Not Rain-Not Rain, Not Rain-Rain-Not Rain, Not Rain-Not Rain-Rain.

If you look into the condition I underscored in quote you should notice that the first two options in a series of events are locked (switched off). Hence we are left only with 0.128 which is 0.384/3 the third option.

I don't understand how you got to the formula 3C1*(0.2)*(0.8)^2=0.384
Can you explain in a little more detail?

rijul007 wrote: I don't understand how you got to the formula 3C1*(0.2)*(0.8)^2=0.384
Can you explain in a little more detail?

stats:binomial probability
n= #trials
k= #successes
n-k= #failures
p= P(success)
q= P(failure)=1-q

nCk * (p^k) * (q^n-k)

n= 3
k= 1
p= 0.2
q= 0.8
3C1*(0.2)*(0.8)^2=0.384
for general reference, you can look up https://www.mathwords.com/b/binomial_pro ... ormula.htm

I don't understand how you got to the formula 3C1*(0.2)*(0.8)^2=0.384
Can you explain in a little more detail?

forget the original question.
If the question was: If the probability of rain on any given day is 0.2 what's the probability that it will rain exactly once in the next three days.

probability of raining any one day (only) in three days = 0.2 * 0.8 * 0.8
Now choose 1 day from the 3 days it can happen = 3C1

probability that it will rain exactly once in the next three days = 3C1 * 0.2 * 0.8 * 0.8 = 3C1*(0.2)*(0.8)^2=0.384