Mathematics: Post your doubts here!

[Anum96]

My question is how do you calculate EFA?

If you remove the lines CB and FE you will get a regular parallelogram. Diagonal inside a parallelogram will bisect the angle and angle EDA will become 70. And so the same diagonal will bisect FAB and make 70. After calculating these angles draw back the lines you omitted EF and CB.Therefore, 360 - 70 - 70 - 120
100.
I hope u get it.

 

[Anum96]

I found it already...thank you missy!

Ohhh

 

[delta.charlie321]

If you remove the lines CB and FE you will get a regular parallelogram. Diagonal inside a parallelogram will bisect the angle and angle EDA will become 70. And so the same diagonal will bisect FAB and make 70. After calculating these angles draw back the lines you omitted EF and CB.Therefore, 360 - 70 - 70 - 120
100.
I hope u get it.

Got it...Thank you

 

[fleurisabelle]

Ohhh

happiness is that I don't need that anymore! only taking AS this session!

 

[awesomaholic101]

Can someone please help with the c part.
TIA.

 

[Eugene99]

Can someone please help with the c part.
TIA.

unable to be read

 

[awesomaholic101]

unable to be read

sorry ... will post a clearer pic

 

[awesomaholic101]

unable to be read

 

[awesomaholic101]

Eugene99
nvm bro ... i got it

 

[husnain99]

Unable to solve last two parts , can someone please explain it ? (Step by step)

 

[Rizwan Javed]

Unable to solve last two parts , can someone please explain it ? (Step by step)

b ii - 56/165?

iii - 1/5?

^ are these the answers? I'll post the solutions if these will be correct.

 

[husnain99]

b ii - 56/165?

iii - 1/5?

^ are these the answers? I'll post the solutions if these will be correct.

Yeah .. Same answers

 

[Rizwan Javed]

Unable to solve last two parts , can someone please explain it ? (Step by step)

The first part says that there shouldn't be any E in three selections of Card.
There are three Cards with Letter E. So the probability of drawing a non-E card ( let it be E' ) for the first time is 8/11.
For the second when he'll pick a card, the probability of drawing E' is 7/10. The sample space has reduced to 10 because the first card picked up has not been replaced.
Third time, the probability of picking E' will be 6/9.

So the probability that no Card with letter E has been picked up in three tries is = 8/11 * 7/10 * 6/9 = 56/165

ii - You can see that there are only two letters that repeat in the word given. 3Es and 2Ns.

First consider that the three cards picked up contain 2Es and one non-E (E'). This can be done in 3 different ways:
E E E' OR E E' E OR E' E E

so the probability that 2Es are there in the three selections is : 3 * 3/11 * 2/10 * 8/9

Now consider that the selections contain 2 Ns and one Non-N (N'). This can also be done in 3 different ways:
N N N' Or N N' N Or N' N N

so the probability that 2Ns are there in the selections is : 3 * 2/11 * 1/10 * 9/9

So the combined probability will be : 3 * 3/11 * 2/10 * 8/9 + 3 * 2/11 * 1/10 * 9/9 = 1/5

OR

more simple it can done in this way: ( 2C2 * 9C1 + 3C2 * 8C1 ) / 11C3 = 1/5

 

[Hassan Ali Abid]

 

[husnain99]

The first part says that there shouldn't be any E in three selections of Card.
There are three Cards with Letter E. So the probability of drawing a non-E card ( let it be E' ) for the first time is 8/11.
For the second when he'll pick a card, the probability of drawing E' is 7/10. The sample space has reduced to 10 because the first card picked up has not been replaced.
Third time, the probability of picking E' will be 6/9.

So the probability that no Card with letter E has been picked up in three tries is = 8/11 * 7/10 * 6/9 = 56/165

ii - You can see that there are only two letters that repeat in the word given. 3Es and 2Ns.

First consider that the three cards picked up contain 2Es and one non-E (E'). This can be done in 3 different ways:
E E E' OR E E' E OR E' E E

so the probability that 2Es are there in the three selections is : 3 * 3/11 * 2/10 * 8/9

Now consider that the selections contain 2 Ns and one Non-N (N'). This can also be done in 3 different ways:
N N N' Or N N' N Or N' N N

so the probability that 2Ns are there in the selections is : 3 * 2/11 * 1/10 * 9/9

So the combined probability will be : 3 * 3/11 * 2/10 * 8/9 + 3 * 2/11 * 1/10 * 9/9 = 1/5

OR

more simple it can done in this way: ( 2C2 * 9C1 + 3C2 * 8C1 ) / 11C3 = 1/5

Alright , can you solve one more question too ? (Last part ) thanks

 

[Rizwan Javed]

Alright , can you solve one more question too ? (Last part ) thanks

Which part? There're two files, so which one?

 

[husnain99]

Which part? There're two files, so which one?

Second one

 

[Rizwan Javed]

Second one

Sorry for getting late :/ I went away for a very important task.

Okay see.

There are 2 chances of going straight to the 100 on the first throw. So the probability of reaching 100 on first throw is 2/36.
There's only 1 way to get a two to go to 99 (probability 1/36) and then back to 96 and then going to 100 (with a probability of 3/36). So the probability of gettin 99 on first throw and then 100 on second throw is 1/36 * 3/36.

Now there're also 33 other options left that make him go beyond 100 on the first throw. So the probablity of getting a 100 on second throw will be : 33/36 * 2/36

the combined probablity wound then be:

2/36 + 1/36 * 3/36 + 33/36 * 2/36 = 47/432

I hope you understood.

 

[Hamighton]

i need a help in Vectors. If you are free then tell me.

 

[husnain99]

Can someone please explain this question ..
Thanks