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mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Guilty and Not Guilty

See table below.

...................Didn't do it........Did it
Guilty................10..................55
Not Guilty.........70..................12

What is the probability of randomly selecting a person (without replacement) who is guilty and then a person who is not guilty from the table?

Note: Guilty person went to jail and this no longer in the table.

Let me see.

A = selecting a person who is guilty.

B = selecting a person who is not Guilty.

Total number of people in the table = 147.

P(A) = 65

P(B) = 82

Considering that one person went to jail means there are now 146 people in the table after event A takes place.

P(A and B) = P(A) • P(B|A)

P(A and B) = 65/147 • 82/146

Is this set up correct?

Keep_Relentless

Member

From: Queensland, Australia

Registered: 2024-05-05

Posts: 61

Re: Guilty and Not Guilty

Yes looks correct.

It's a bit under 1/4.

I think you understand how to do these problems

---

"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Guilty and Not Guilty

Yes looks correct.

It's a bit under 1/4.

I think you understand how to do these problems

I am slightly lost in terms of definitions.  Can you please explain independent and dependent events?

Can you please explain with replacement and without replacement?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Guilty and Not Guilty

independant/dependent

Suppose I have a box (1) of coloured marbles; 5 red and 3 blue; and another box (2) with 4 red and 4 blue.

I'm going to choose a box (at random) and then a marble (without looking). What's the probability it's red.

The chances of getting a red will be different if I choose one box from what it will be if I chose the other.

We say P(red) is dependent on the box I choose.

If box 1 then P(red) = 5/8
If box 2 then P(red) = 4/8   (I won't bother with simplifying at this stage)

I'm choosing which box at random so P(box 1) = 1/2 = P(box 2)

So P(red) = P(box 1) x P(red if that box) + P(box 2) x P(red if that box) = 1/2 x 5/8 + 1/2 x 4/8 = 5/16 + 4/16 = 9/16
(note it was easier to do the adding when I hadn't simplified the fractions to their lowest terms)

P(blue) =  P(box 1) x P(blue if that box) + P(box 2) x P(blue if that box) = 1/2 x 3/8 + 1/2 x 4/8 = 3/16 + 4/16 = 7/16

Useful extra check: P(red) + P(blue) = 9/16 + 7/16 = 16/16 = 1. as it must be because one of those two possibilities must happen.

Now for independent.

I have a dice (usual 6 faces numbered 1-6) and a pack of cards (usual 52 ace to king four suits)

I throw the dice and pick a card at random. What's the probability of getting a six on the dice and an ace on the card?

These are independent events. What the dice does when I throw it has no effect of which card I pick.

P(six and ace) = 1/6 x 4/52 = 1/6 x 1/13 = 1/78

Now for replacement/ no replacement

Forget the dice and just use the pack of cards.

I pick two cards at random. What's the probability of getting 2 aces.

If I want to end up holding both cards then I must use no replacement. Otherwise I won't end up with 2 cards.

P(first ace) = 4/52     P(second ace) = 3/51       P(two aces) = 4/52 x 3/51 = 1/13 x 1/17 = 1/221

Each day I shuffle the 52 cards and pick one. What's the probability of getting 2 aces on two consecutive days?

This time there is replacement because it says I start each day with 52 cards. P(2 aces) = 4/52 x 4/52 = 1.13 x 1/13 = 1/169

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Guilty and Not Guilty

independant/dependent

Suppose I have a box (1) of coloured marbles; 5 red and 3 blue; and another box (2) with 4 red and 4 blue.

I'm going to choose a box (at random) and then a marble (without looking). What's the probability it's red.

The chances of getting a red will be different if I choose one box from what it will be if I chose the other.

We say P(red) is dependent on the box I choose.

If box 1 then P(red) = 5/8
If box 2 then P(red) = 4/8   (I won't bother with simplifying at this stage)

I'm choosing which box at random so P(box 1) = 1/2 = P(box 2)

So P(red) = P(box 1) x P(red if that box) + P(box 2) x P(red if that box) = 1/2 x 5/8 + 1/2 x 4/8 = 5/16 + 4/16 = 9/16
(note it was easier to do the adding when I hadn't simplified the fractions to their lowest terms)

P(blue) =  P(box 1) x P(blue if that box) + P(box 2) x P(blue if that box) = 1/2 x 3/8 + 1/2 x 4/8 = 3/16 + 4/16 = 7/16

Useful extra check: P(red) + P(blue) = 9/16 + 7/16 = 16/16 = 1. as it must be because one of those two possibilities must happen.

Now for independent.

I have a dice (usual 6 faces numbered 1-6) and a pack of cards (usual 52 ace to king four suits)

I throw the dice and pick a card at random. What's the probability of getting a six on the dice and an ace on the card?

These are independent events. What the dice does when I throw it has no effect of which card I pick.

P(six and ace) = 1/6 x 4/52 = 1/6 x 1/13 = 1/78

Now for replacement/ no replacement

Forget the dice and just use the pack of cards.

I pick two cards at random. What's the probability of getting 2 aces.

If I want to end up holding both cards then I must use no replacement. Otherwise I won't end up with 2 cards.

P(first ace) = 4/52     P(second ace) = 3/51       P(two aces) = 4/52 x 3/51 = 1/13 x 1/17 = 1/221

Each day I shuffle the 52 cards and pick one. What's the probability of getting 2 aces on two consecutive days?

This time there is replacement because it says I start each day with 52 cards. P(2 aces) = 4/52 x 4/52 = 1.13 x 1/13 = 1/169

Bob

Wow!! You outdid yourself here. What will be my next excuse?

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Guilty and Not Guilty

independant/dependent

Suppose I have a box (1) of coloured marbles; 5 red and 3 blue; and another box (2) with 4 red and 4 blue.

I'm going to choose a box (at random) and then a marble (without looking). What's the probability it's red.

The chances of getting a red will be different if I choose one box from what it will be if I chose the other.

We say P(red) is dependent on the box I choose.

If box 1 then P(red) = 5/8
If box 2 then P(red) = 4/8   (I won't bother with simplifying at this stage)

I'm choosing which box at random so P(box 1) = 1/2 = P(box 2)

So P(red) = P(box 1) x P(red if that box) + P(box 2) x P(red if that box) = 1/2 x 5/8 + 1/2 x 4/8 = 5/16 + 4/16 = 9/16
(note it was easier to do the adding when I hadn't simplified the fractions to their lowest terms)

P(blue) =  P(box 1) x P(blue if that box) + P(box 2) x P(blue if that box) = 1/2 x 3/8 + 1/2 x 4/8 = 3/16 + 4/16 = 7/16

Useful extra check: P(red) + P(blue) = 9/16 + 7/16 = 16/16 = 1. as it must be because one of those two possibilities must happen.

Now for independent.

I have a dice (usual 6 faces numbered 1-6) and a pack of cards (usual 52 ace to king four suits)

I throw the dice and pick a card at random. What's the probability of getting a six on the dice and an ace on the card?

These are independent events. What the dice does when I throw it has no effect of which card I pick.

P(six and ace) = 1/6 x 4/52 = 1/6 x 1/13 = 1/78

Now for replacement/ no replacement

Forget the dice and just use the pack of cards.

I pick two cards at random. What's the probability of getting 2 aces.

If I want to end up holding both cards then I must use no replacement. Otherwise I won't end up with 2 cards.

P(first ace) = 4/52     P(second ace) = 3/51       P(two aces) = 4/52 x 3/51 = 1/13 x 1/17 = 1/221

Each day I shuffle the 52 cards and pick one. What's the probability of getting 2 aces on two consecutive days?

This time there is replacement because it says I start each day with 52 cards. P(2 aces) = 4/52 x 4/52 = 1.13 x 1/13 = 1/169

Bob

Look what I found on this site:

Replacement

Note: if we replace the marbles in the bag each time, then the chances do not change and the events are independent:

With Replacement: the events are Independent (the chances don't change)
Without Replacement: the events are Dependent (the chances change)

Cook, right?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Guilty and Not Guilty

That sums it up nicely.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Guilty and Not Guilty

That sums it up nicely.

Bob

I am going to play with conditional probability questions like this today and probably tomorrow. No need to rush into the next topic. By the way, I will return to my college algebra textbook some time this week.