Competition - post a photo you've taken of nature >>

Probability help please

Really struggling - keep thinking I've cracked probability then i find a question I cant answer.
I dont know where to begin with this one:

At the end of a training course students must pass a test to gain a diploma. The probability of passing first time is 0.8.
those who fail are allowed one more attempt. The probability of of passing the resit is 0.65.
Two friends Sam and Tim follow the course and take the test.
What is the probability they both gain a diploma?

I really cant think where to start with this. I can't make a tree diagram that works!! Not sure how to approach it. Could anybody help me please?

Reply 1

Zacken

Original post by JayD29

What d'you mean by not being able to make a tree diagram?

The first branch should be "passes" and "not passes" with probability 0.8 and 0.2 respectively and then the passes branch has nothing more after it but the "not passes" branch splits into two other branches "passes resit" and "not passes resit" with probability 0.65 and 1-0.65 respectively.

Reply 2

ghostwalker

Original post by JayD29

Do a tree diagram for just one person.

Then what's the probability that Sam passes?

Similarly Tim.

Then assuming they're independent, probability they both pass is?

Reply 3

GUMI

Like everyone has said, draw a tree diagram then draw the probability of that event happening for each branch. To find the final probability for each outcome, follow the branch and multiply the probabilities

Reply 4

JayD29

Oh man...can't believe how simple that was. I just couldn't figure out where to start!! All done now. Thank you so much.

I have one more question that I can't get my head around - if I post it I wonder of one of you guys could maybe help me with it please?

Bag A contains 5 red and 3 blue counters. Bag B contains 2 red and 6 blue counters.
Step 1: a counter is taken from bag a and placed in bag b.
Step 2: a counter is taken from bag b and placed in bag a.
Calculate the probability that Bag a has more red counters than blue counters after these 2 steps.

There's just something about the way these questions are worded that confuses me - I just dont know where to start.

Reply 5

ghostwalker

Original post by JayD29

Again a tree diagram will cover it.

We can write the number of red and number of blue in bag A as an order pair (#red, #blue)

So we start at (5,3)
Assuming counters are taken at random, we take a red counter with probability 5/(5+3), and this moves us to position (4,3). Similarly blue.

Can you carry on and complete the tree now?

Reply 6

JayD29

Original post by ghostwalker

I've tried making a tree starting with bag A having (5, 3) but then I dont know how to move onto bag b.
I said that we started with Bag a
5/8 red
3/8 blue
this then moves onto bag b which will be either 3/9 red and 6/9 blue or 2/9 red and 7/9 blue depeding on what is chosen from bag a. But then I dont know how to move back to bag a, and suspect I'm overcomplicating things again....

Reply 7

JayD29

Could anyone point me in the right direction with this please?

Bag A contains 5 red and 3 blue counters. Bag B contains 2 red and 6 blue counters.
Step 1: a counter is taken from bag A and placed in bag B.
Step 2: a counter is taken from bag B and placed in bag A.
Calculate the probability that Bag A has more red counters than blue counters after these 2 steps.

There's just something about the way these questions are worded that confuses me - I just dont know where to start.
I've tried to draw a tree diagram but after the intitial 5/8 and 3/8 Im not sure where to go with it. I moved onto bag B then having either 3/9 and 6/9 or 2/9 and 7/9 but then dont know where to go from there...or even if that was the right thing to do!

Reply 8

Farhan.Hanif93

Original post by JayD29

By drawing a tree diagram, you've noticed that there are a multiple ways to ensure that bag A has more red counters than blue counters - and, therefore, that calculating the probability this way is likely to be messy (although, still reasonably easy to calculate if you think of the possibilities individually and work them through one-by-one).

Instead, consider the complementary event (-often a good idea-) that bag A has at least as many blue counters as there are red counters. There is only one way for this event to occur, what is it? And what is the probability of this event occurring? Then subtracting this probability from 1 will yield the probability that you were after in the first place.

(edited 7 years ago)

Reply 9

ghostwalker

Original post by JayD29

See picture. Ignore the attached thumbnails - can't get the editor to remove them.

Step1.

Suppose we choose a red, then we go from (5,3) to (4,3) with probability 5/8.

Now bag B will have (3,6). You could record that on your diagram for reference if you wish.

Then step 2.

Suppose we chose red again (from bag B this time), so bag A goes from (4,3) to (5,3) with probability 3/9.

And repeat for the other options