Binomial distribution

• Apr 25th 2010, 12:03 AM
Coach
Binomial distribution
Hello!

My Problem:
A crossword puzzle is published in a magazine 6 times a week(all days except Sunday). A woman is able to complete on average eight out of ten crosswords. Given that she completes the puzzle on Monday, find the probability that she will complete at least four in the rest of the week.

My solution:
Let the random variable $\displaystyle X$ denote the number of crossword puzzles completed. The probability that she solves a crossword puzzle on Monday is $\displaystyle 0.8$.
Hence for the rest of the week
X follows a binomial distribution with parameters (5,0.8).
So
$\displaystyle P(X \geq 4| X=1)=\frac{0.737}{0.8}=0.922$

and it is wrong. The right answer according to the book is $\displaystyle 0.737$.

Any help is appreciated.
• Apr 25th 2010, 01:29 AM
losm1
It looks to me like you don't have to divide out 0.8 - just basic cumulative probability will be enough:

$\displaystyle P(X\ge4)=P(4)+P(5)=\left({5 \choose 4} \times 0.8^4\times 0.2^1\right) + 0.8^5 = 0.737$
• Apr 25th 2010, 10:08 PM
acc100jt
Note that the events are independent, whether she complete the puzzle on Monday will not affect the result for the remaining days.

So $\displaystyle P(X \ge 4 | puzzle \; completed \;on\; Monday) = P(X \ge 4)=P(X=4)+P(X=5)$