Simple Probability Question (1 Viewer)

Ambility · Sep 30, 2015

I struggle with this topic, though I'm trying to get better. I'd appreciate it if you could show me the working for this question:

$Sophia, Gabriel and Elizabeth take their driving test. The chances they pass are $\frac{1}{2}$, $\frac{5}{8}$, and $\frac{3}{4}$ respectively. If only one of them passes, find the probability it is Gabriel.$\\\\$How many times would you need to roll a die so the probability of tossing rolling at least one six is greater than $\frac{9}{10}$?$

Drsoccerball · Sep 30, 2015

Ambility said:
I struggle with this topic, though I'm trying to get better. I'd appreciate it if you could show me the working for this question:

$Sophia, Gabriel and Elizabeth take their driving test. The chances they pass are $\frac{1}{2}$, $\frac{5}{8}$, and $\frac{3}{4}$ respectively. If only one of them passes, find the probability it is Gabriel.$

$\frac{5}{8} \cdot \frac{1}{2} \cdot \frac{1}{4}$

So we multiply the events of G passing and S and E failing.

davidgoes4wce · Sep 30, 2015

Think of the probability of Sophia and Elizabeth failing as the complement (1-1/2=1/2 for Sophia) and (1-3/4=1/4 for Elizabeth)

Ambility · Sep 30, 2015

Drsoccerball said:
$\frac{5}{8} \cdot \frac{1}{2} \cdot \frac{1}{4}$

So we multiply the events of G passing and S and E failing.

The answer in the back of the book is $\frac{5}{17}$ . Isn't the sample space, or the number of possible outcomes reduced when one person, and only one, must pass? Question 16c, 10C Cambridge 3U HSC for reference.

Drsoccerball · Sep 30, 2015

Ambility said:
The answer in the back of the book is $\frac{5}{17}$ . Isn't the sample space, or the number of possible outcomes reduced when one person, and only one, must pass? Question 16c, 10C Cambridge 3U HSC for reference.

No because the probability of one person passing doesn't depend on the others results

Ambility · Sep 30, 2015

Drsoccerball said:
No because the probability of one person passing doesn't depend on the others results

But the question did not say at least one person passed, it said only one person passed. This means that S,G passing but E failing is not a desirable outcome. Are you sure this doesn't effect the answer? Is the textbook answer incorrect?

Drsoccerball · Sep 30, 2015

Ambility said:
But the question did not say at least one person passed, it said only one person passed. This means that S,G passing but E failing is not a desirable outcome. Are you sure this doesn't effect the answer? Is the textbook answer incorrect?

Ambility said:
I struggle with this topic, though I'm trying to get better. I'd appreciate it if you could show me the working for this question:

$$Sophia, Gabriel and Elizabeth take their driving test. The chances they pass are $\frac{1}{2}$, $\frac{5}{8}$, and $\frac{3}{4}$ respectively$ . If only one of them passes, $find the probability it is Gabriel.$\\\\$How many times would you need to roll a die so the probability of tossing rolling at least one six is greater than $\frac{9}{10}$?$$

He said only one passes.
.

rand_althor · Sep 30, 2015

Isn't it a conditional probability problem?

Drsoccerball · Sep 30, 2015

rand_althor said:
Isn't it a conditional probability problem?

The question said only one passes so theres no conditions.

rand_althor · Sep 30, 2015

Drsoccerball said:
The question said only one passes so theres no conditions.

Oh. I thought the condition would be only one passing.

Ambility · Sep 30, 2015

rand_althor said:
Oh. I thought the condition would be only one passing.

I thought so too.

rand_althor · Sep 30, 2015

$$P(only Sophia passes) $=\frac{1}{2} \times \frac{3}{8} \times \frac{1}{4}=\frac{3}{64}$. P(only Gabriel passes) $=\frac{1}{2} \times \frac{5}{8} \times \frac{1}{4}=\frac{5}{64}$. P(only Elizabeth passes) $=\frac{1}{2} \times \frac{3}{8} \times \frac{3}{4}=\frac{9}{64}$. So P(only one passes)$=\frac{3+5+9}{64}=\frac{17}{64}$. So the probability that if only one passes, it is Gabriel is $\frac{\frac{1}{2} \times \frac{5}{8} \times \frac{1}{4}}{\frac{17}{64}}=\frac{\frac{5}{64}}{\frac{17}{64}}=\frac{5}{17}$.$

Someone correct me if I'm wrong.

Ekman · Sep 30, 2015

rand_althor said:
$$P(only Sophia passes) $=\frac{1}{2} \times \frac{3}{8} \times \frac{1}{4}=\frac{3}{64}$. P(only Gabriel passes) $=\frac{1}{2} \times \frac{5}{8} \times \frac{1}{4}=\frac{5}{64}$. P(only Elizabeth passes) $=\frac{1}{2} \times \frac{3}{8} \times \frac{3}{4}=\frac{9}{64}$. So P(only one passes)$=\frac{3+5+9}{64}=\frac{17}{64}$. So the probability that if only one passes, it is Gabriel is $\frac{\frac{1}{2} \times \frac{5}{8} \times \frac{1}{4}}{\frac{17}{64}}=\frac{\frac{5}{64}}{\frac{17}{64}}=\frac{5}{17}$.$

Someone correct me if I'm wrong.

I did that also

davidgoes4wce · Sep 30, 2015

OK I missed the 'if' part to this question

InteGrand · Sep 30, 2015

Yes, the driving test question is a conditional probability question.

braintic · Sep 30, 2015

Confirming ... the answer IS 5/17.

flavurr · Sep 30, 2015

Did you only need help with the driving q?

Ambility · Sep 30, 2015

flavurr said:
Did you only need help with the driving q?

I also need help with this one:

$How many times would you need to roll a die so the probability of tossing rolling at least one six is greater than $\frac{9}{10}$?$

InteGrand · Sep 30, 2015

Ambility said:
I also need help with this one:

$How many times would you need to roll a die so the probability of tossing rolling at least one six is greater than $\frac{9}{10}$?$

$The probability of rolling a six on a given roll is $p=\frac{1}{6}$, so the probability of failing to roll a six on a given roll is $q=\frac{5}{6}$. So in $n$ rolls ($n\in \mathbb{Z}^+$), the probability of rolling at least one six is $1-\Pr(\text{no sixes})=1-q^n = 1-\left(\frac{5}{6} \right)^n$. So we just need to solve $1-\left(\frac{5}{6} \right)^n >\frac{9}{10}$, which I assume you know how to do (take the smallest integer $n$ that satisfies this inequation to get the minimal number of required rolls).$

Simple Probability Question (1 Viewer)

Ambility

Active Member

Drsoccerball

Well-Known Member

davidgoes4wce

Well-Known Member

Ambility

Active Member

Drsoccerball

Well-Known Member

Ambility

Active Member

Drsoccerball

Well-Known Member

rand_althor

Active Member

Drsoccerball

Well-Known Member

rand_althor

Active Member

Ambility

Active Member

rand_althor

Active Member

Ekman

Well-Known Member

davidgoes4wce

Well-Known Member

InteGrand

Well-Known Member

braintic

Well-Known Member

flavurr

New Member

Ambility

Active Member

InteGrand

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)