1. ## probability

Two questions Herman and Amos are usually late for school. Herman travels to school either by train, bus or taxi with the probability of 2/5, 2/5 and 1/5 respectively. His classmate Amos will only travel by bus or taxi with probability of 3/4 for travelling by bus. Herman has the probability 1/4, 1/2 and 1/4 of being late when he travels by the train, bus or taxi respectively. Amos has the probability of 3/5 and 1/5 of being late when he travels by bus or taxi respectively.

Find the probability that

i) Either of them is late

ii) they used the same mode of transport given that both of them are late today.

answer to i) is 27/40 and ii) is 37/40 but i've tried for ages and can't get the answer! (It's not the kind of question i'm used to - there seem to be two separate probability trees here? It's strange.) please help. thanks.

2. ## Re: probability

Hello, thesocialnetwork!

Here's part (a) . . .

Herman and Amos are usually late for school.
Herman travels to school either by train, bus or taxi with the probability of 2/5, 2/5 and 1/5 respectively.
His classmate Amos will only travel by bus or taxi with probability of 3/4 for travelling by bus.
Herman has the probability 1/4, 1/2 and 1/4 of being late when he travels by the train, bus or taxi respectively.
Amos has the probability of 3/5 and 1/5 of being late when he travels by bus or taxi respectively.

Find the probability that: (a) Either of them is late.

$\displaystyle \text{Let: }\:\begin{Bmatrix}H &=& \text{Herman is late} \\ A &=& \text{Amos is late} \end{array}$

We want: .$\displaystyle P(H\vee A) \;=\;P(H) + P(A) - P(H\wedge A)$ .[1]

Herman

$\displaystyle \begin{array}{cccc cccc ccccc}P(\text{train}) &=& \frac{2}{5} && P(\text{late}) &=& \frac{1}{4} & \Rightarrow &P(\text{train} \wedge \text{late}) &=& \frac{1}{10} \\ \\[-4mm]P(\text{bus}) &=& \frac{2}{5} && P(\text{late}) &=& \frac{1}{2} & \Rightarrow& P(\text{bus} \wedge \text{late}) &=& \frac{1}{5} \\ \\[-4mm]P(\text{taxi}) &=& \frac{1}{5} && P(\text{late}) &=&\frac{1}{4} & \Rightarrow& P(\text{taxi}\wedge\text{late}) &=& \frac{1}{20} \end{array}$

.$\displaystyle P(H) \;=\;\tfrac{1}{10} + \tfrac{1}{5} + \tfrac{1}{20} \;=\;\tfrac{7}{20}$

Amos

$\displaystyle \begin{array}{cccc cccc ccccc}P(\text{bus}) &=& \frac{3}{4} && P(\text{late}) &=& \frac{3}{5} &\Rightarrow& P(\text{bus}\wedge\text{late}) &=& \frac{9}{20} \\ \\[-4mm] P(\text{taxi}) &=& \frac{1}{4} && P(\text{late}) &=& \frac{1}{5} &\Rightarrow& P(\text{taxi}\wedge\text{late}) &=& \frac{1}{20}\end{array}$

. . $\displaystyle P(A) \;=\;\tfrac{9}{20} + \tfrac{1}{20} \;=\;\tfrac{1}{2}$

Hence: .$\displaystyle P(H\wedge A) \;=\;\tfrac{7}{20}\cdot\tfrac{1}{2} \;=\;\tfrac{7}{40}$

Substitute into [1]:
. . . . $\displaystyle P(H \vee A) \;=\;\frac{7}{20} + \frac{1}{2} - \frac{7}{40} \;=\;\frac{27}{40}$

3. ## Re: probability

thanks so much!

4. ## Re: probability

can you get part ii) though? I got 5/7 which is wrong apparently

5. ## Re: probability

Hello again , thesocialnetwork!

I don't agree with their answer.
Please check my reasoning and my math.

(b) They used the same mode of transport, given that both of them were late.

We want: .$\displaystyle P(\text{same mode}\,|\,\text{both late}) \;=\;\frac{P(\text{same mode} \wedge\text{both late})}{P(\text{both late})}$ . [1]

$\displaystyle \text{Let: }\:\begin{Bmatrix}H &=& \text{Herman is late} \\ A &=& \text{Amos is late} \end{array}$

Herman

$\displaystyle \begin{array}{cccc cccc ccccc}P(\text{train}) &=& \frac{2}{5} && P(\text{late}) &=& \frac{1}{4} & \Rightarrow &P(\text{train} \wedge \text{late}) &=& \frac{1}{10} \\ \\[-4mm]P(\text{bus}) &=& \frac{2}{5} && P(\text{late}) &=& \frac{1}{2} & \Rightarrow& P(\text{bus} \wedge \text{late}) &=& \frac{1}{5} \\ \\[-4mm]P(\text{taxi}) &=& \frac{1}{5} && P(\text{late}) &=&\frac{1}{4} & \Rightarrow& P(\text{taxi}\wedge\text{late}) &=& \frac{1}{20} \end{array}$

.$\displaystyle P(H) \;=\;\tfrac{1}{10} + \tfrac{1}{5} + \tfrac{1}{20} \;=\;\tfrac{7}{20}$

Amos

$\displaystyle \begin{array}{cccc cccc ccccc}P(\text{bus}) &=& \frac{3}{4} && P(\text{late}) &=& \frac{3}{5} &\Rightarrow& P(\text{bus}\wedge\text{late}) &=& \frac{9}{20} \\ \\[-4mm] P(\text{taxi}) &=& \frac{1}{4} && P(\text{late}) &=& \frac{1}{5} &\Rightarrow& P(\text{taxi}\wedge\text{late}) &=& \frac{1}{20}\end{array}$

. . $\displaystyle P(A) \;=\;\tfrac{9}{20} + \tfrac{1}{20} \;=\;\tfrac{1}{2}$

$\displaystyle \text{Denominator: }\:P(\text{both late}) \;=\;P(H\wedge A) \;=\;\frac{7}{20}\cdot\frac{1}{2} \;=\;\frac{7}{40}$

$\displaystyle P(\text{both: bus }\wedge\text{ late}) \:=\:\tfrac{1}{5}\cdot\tfrac{9}{20} \:=\:\tfrac{9}{100}$

$\displaystyle P(\text{both: taxi }\wedge\text{ late}) \:=\:\tfrac{1}{20}\cdot\tfrac{1}{20} \:=\:\tfrac{1}{400}$

$\displaystyle \text{Numerator: }\:P(\text{same mode }\wedge\text{ both late}) \:=\:\frac{9}{100} + \frac{1}{400} \:=\:\frac{37}{400}$

Substitute into [1]:
. . . $\displaystyle P(\text{same mode}\,|\,\text{both late}) \;=\;\dfrac{\frac{37}{400}}{\frac{7}{40}} \;=\;\frac{37}{70}$

6. ## Re: probability

I agree completely. I am going to go with yours. Thank you very much; there is great clarity in your response. (: