A certain cancer treatment successfully treats the cancer 70% each time it is administered. My mom survived for 8 months using t
1 answer:
Answer:
5.80%
Step-by-step explanation:
Given:
The probability of successful treatment of cancer is,
The probability of successful treatment in a given month is independent of the other month. So, probability of successful treatment for 'n' months in a row is given as:
Plug in 8 for 'n' and determine the required probability. This gives,
Therefore, the probability that the treatment was successful 8 months in a row is 5.80% rounded to nearest tenth of a percent.
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Answer:
144
Step-by-step explanation:
For a bitstring of length n, there are Fibonacci(n+2) strings containing no two consecutive zeros. This can be seen by constructing the strings starting with n=1.
1-bit strings: 1, 0 -- 2 strings not containing consecutive 0s
2-bit strings: 11, 10, 01 -- 3 strings not containing consecutive 0s
Note that we have added 1 to all the 1-bit strings, and added 0 only to the string ending in 1.
3-bit strings: 111, 110, 101, 011, 010 -- 5 strings not containing consecutive 0s
Note that these 5 strings consist of all (3) of the 2-bit strings with 1 appended, and all (1) of the 2-bit strings ending in 1 with 0 appended. The number that now end in 0 is the number previously ending in 1.
__
If (x, y) represents the numbers of n-bit strings ending in (0, 1), then the number of (n+1)-bit strings ending in (0, 1) is (y, x+y). That is, the recursive relation is ...
For n=1 to n=10, these pairs are ...
(1, 1), (1, 2), (2, 3), (3, 5), (5, 8), (8, 13), (13, 21), (21, 34), (34, 55), (55, 89)
The sequence of b[n] values is ...
2, 3, 5, 8, 13, 21, 34, 55, 89, 144
which are the n=3 to n=12 numbers from the Fibonacci sequence.
That is, there will be Fibonacci(12) = 144 10-bit strings with no consecutive 0s.
Given:
Total number of students = 27
Students who play basketball = 7
Student who play baseball = 18
Students who play neither sports = 7
To find:
The probability the student chosen at randomly from the class plays both basketball and base ball.
Solution:
Let the following events,
A : Student plays basketball
B : Student plays baseball
U : Union set or all students.
Then according to given information,
We know that,
Now,
It means, the number of students who play both sports is 5.
The probability the student chosen at randomly from the class plays both basketball and base ball is
Therefore, the required probability is .