A Probability Problem

Dennis started reviewing a week before his exam. The probability of him passing the test given that he did not review at all is $0.3$ , and that chance increases by $0.1$ for each day that he spends reviewing (maximum of $6$ days). If on any given day, he has a even chance between reviewing and doing something else, what is the probability of passing the exam?

(Note: This problem was taken from my math statistics exam.)

#Combinatorics #Probability

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Make cases. $\begin{aligned} & 1) \ \text{He doesn't review at all and passes} \\ & 2) \ \text{He reviews one day and passes}\\ & \vdots\\ &\vdots \end{aligned}$

This leads us to the following

$P(\text{Dennis passes})=\dfrac{1}{640}\left(\displaystyle\sum_{n=0}^{6}\left[ \binom{6}{n} (3+n)\right]\right)=\dfrac{3}{5}=0.6$

Pratik Shastri - 6 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$