Math Is Fun Forum

There are two ways of getting a probability: theoretical and experimental.

eg. Theoretical: What's the probability of drawing an ace from a pack of cards. We can ask

In this case, there are 52 events of which 4 give us an ace (but check they're equally likely) so 4/52

eg Experimental.  I've got a box of 100 similar drawing pins. I give it a good shake and scatter them on a table surface. Some may land pin up and some may land pin touching the table. Calculate the P(pin up) by counting

I doubt you could get an answer any other way but by doing the experiment.  If a pin were made with a 1cm diameter base and a 1metre pin then I'm pretty confident P(pin up) would be zero. If the base were large, say 1 metre and the pin tiny, say 1 mm. then I think they'd all land pin up. But for normal pins there's no way to theoretically assess this.

So back to the question.  Is there any theory that could help here? Don't think so. Could we use an experiment? Only if we have a large number of parallel universes where the event has already happened.  So P(Jesus returns today) is indeterminant.  I also think that's the point. Jesus didn't want his followers to cram a load of good works the day before His return. He wanted them to be ever ready; ie. behave as instructed all the time every day. The Koran carries a similar message.

Bob

The standard general equation for a circle is derived like this.

If the centre is at (a,b) and rhe radius is r and (x,y) is the general point on the circumference:

Horizontal distance from the centre to the point is x-a
Vertical distance from the centre to the point is y-b

There's a right angle between the horizontal and vertical lines so Pythagoras can be applied

You can use this for these questions:

1. You're told a and b and you can use the distance from the centre to the given point as the radius.

2.  The tangent must be the x axis so at point (3,0) .  The radius must be 5.  So, once again, we have the radius and the centre.

3. Similar but now its the y axis.

4. Calculate the distance between (3,5) and (0,0). This is the radius.

5. There's more to do here. You can calculate the length of the diameter and hence get the radius. Also the midpoint of the diameter will be the centre. So that's enough to get the equation.

The y intercepts will have x coordinate zero so substitute x=0 in the equation and solve for y (two solutions).

Bob

I'll show one then hopefully you can do the rest.

1.  |X-1| < 1/2

Bob

Parts 1 and 2 correct.

You've done 2 yellows. Find also 2 reds, then add the two  answers to get P(two same colour)

Last part: Add P(red then yellow) to P(yellow then red)

Bob

(x+a)^2= x^2 + 2ax + a^2

The 2ax term is missing . Only explanation is a=0. Similarly for 2by.

Bob

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

Thanks for pointing me to this effect. I get it too. But why?  Neither is what I wanted.

It is something that didn't happen when I first joined (2010).  I suspect it happened when the server was changed.  Before then a poster could upload images too.

Bob

The frac{}{} is missing. Then the expression needs to be in {} after the square root command.

Did you still want ^5 as well. If yes then you could do this ^{\frac{5}{2}}

When I first joined the forum a command square brackets code could surround the math commands to stop the interpreter activating the commands. This allowed a poster to show a Latex command without it actually happening. All I see now is a black rectangle.  I'll give you an example:

What do others see?

Bob

Spam post removed. Unfortunately a novice had added two new, genuine posts to the thread.  When the lead post is deleted everything that follows goes too, so I copied the content into a new Help Me post. Best way to let me know when you spot any spam is to make a report.[link at bottom of the post] All mods and admin get your report when they next log in and so the first can delete the offending post and 'tick off' the report as dealt with.  That's quicker and safer than making a comment in a different post as (1) another mod might have been able to deal with it more quickly and (2) I might not even look at your post as it might not be one for me.

Using Excel I have found that the third number is divisible by 29 and 13 (but not 7 and 5 also not 13^2 nor 29^2)

That makes it a bit easier to find any other factors.

I'll keep looking.

Q2. I recently made a post for someone else about the fibonacci sequence and the golden ratio.  I set some hw but no one, including the OP, has responded.  But I did discover a formula for fibonacci numbers in Wiki and I think this could be used to solve this one. It won't be simple

I have done what I said with respect to your old account. You should have received an email from Matilda Strong.  Did you spot it?

Bob

Correct

B

It ought to say "If Janet choses TRUE or FALSE at random"

Bob