Answer:
The answer is c
Step-by-step explanation:
If an event occurs 0 times (out of 4, in this case) then it does not occur at least once. So we can find the probability of it not occurring and then subtract that value from 1.
So, what are the chances of it not occurring in 1 trip?
1−.37=.63
What about not occurring in 2 trips?
(1−.37)⋅(1−.37)=.63⋅.63=.3969
Now what about not occurring in 4 trips?
.63^4 = 0.15752961
We must subtract this value from 1
(recall that what we just calculated is the probability of it not occurring, so the probability of it occurring at least once is:
1−0.15752961 = .84247039
TLDR - In 4 trips the chance of a guest catching a cutthroat once in under 4 trips in 0.84.
Hi,
f(1)=16
f(2)=f(1)+2*2=16+4=20
f(3)=f(2)+2*3=20+6=26
f(4)=f(3)+2*4=26+8=34
Are there any other details as to what the numbers are? otherwise there are six possibilities for this
12+54=66
66=6(_+_)
11=(_+_)
1) 0, 11
2) 1, 10
3) 2, 9
4) 3, 8
5) 4, 7
6) 5, 6
