Please see the attachment and answer with explanation.

- anonymous

Please see the attachment and answer with explanation.

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- schrodinger

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- anonymous

Please see these and answer.

##### 1 Attachment

- anonymous

der?

- dumbcow

haha thats like a whole assignment on probability
please post one question at a time

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## More answers

- anonymous

lol..yeahh..but they r tiny ones...

- anonymous

At least solve questions 1, 2 , 3.

- vishal_kothari

the link is not opening in my computer..so draw the questions...

- vishal_kothari

draw it...

- anonymous

wait.

- anonymous

Question 1 ) From a group of 3 boys and 2 girls, 2 children are selected at random. Find the Sample Space.

- vishal_kothari

options...

- anonymous

Options nahin hain. Model paper for SA-2 question.

- anonymous

Question 2) numbers are thrown getting 2 numbers whose sum is divisible by 4 or 5, is considered as a success. Find the probability og getting success.

- vishal_kothari

what is tough in that...

- anonymous

Question1) answer is 20. I guess.

- anonymous

Question 3) A coin is tossed 3 times. Find the chance that head and tail show alternately. (I think answer is 4/8 = 1/2) Plz verify and say.

- anonymous

Ok, I'll take a crack at this for you.
So, for #1, the Sample Space is defined as the list of all possible outcomes. The question is, is 1 boy and 1 girl the option? Or are they unique, like Boy 1 and Girl 1 vs. Boy 1 and Girl 2 - does that make sense? So, I'm not 100% on how the question is defined to be honest...
#2: You need to determine all the possible combinations of the sum of two die (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12), see the ones divisible by 4 or 5, then divide that number by they number of possibilities (in this example there are 11).
#3: A coin flip is a 50/50 chance. Alternatively, this written as 1/2 (1 possible successful outcome out of two possible outcomes). So, when flipping a coin three times, you need to multiply the respective probabilities to find your answer.
Hope this helps

- anonymous

Can u please explain questions 1 and 3 in detail? The detailed solution.

- anonymous

For #1 I really can't - the problem doesn't give enough parameters for me to be sure. The issue is how they define the Sample Space. If one boy and one girl is a possible outcome, is it just ONE possible outcome? Does it matter WHICH boy or girl are chosen? Also, does the order matter? Is one boy and one girl the same or different from one girl and one boy?
#3 What exactly are you looking for here? Which part of the explanation do you need help with?

- anonymous

In question 1, waht matters is the grouping I think. Means Boy-boy, boy0girl, and so on.
In question 3, The sample space is: (HHH), (HHT), (HTH), (HTT), (TTT), (TTH), (THT) and (THH). So I think probability of getting head and tail alternately is 4/8 = 1/2.

- anonymous

For #1, ok, so in that case it would be 20 then?
For #3, aren't there only two correct outcomes? THT & HTH?

- anonymous

THen ok. For no.3, I am doubt. I think u r correct.

- anonymous

And for #1 - my concern is this, here are the two sample spaces which I can't decide between:
BG, BB, GG, GB
OR
B1G1, B1G2, B2G1, B2G2, B3G1, B3G2, and so on...

- anonymous

I tried with first assumption though I am not sure.

- anonymous

Ok sweet, FFM is here - I really want to see his opinion on this

- anonymous

yeah sure.

- anonymous

>> From a group of 3 boys and 2 girls, 2 children are selected at random. Find the Sample Space.
I will just answer the number in the sample space: \( \binom 32 +\binom 22+ \binom 31\times \binom21 =\cdots\)

- anonymous

So the final answer? I don't know Permutation and combination those things.

- anonymous

How could you possible understand probability without understanding elementary combinatorics?

- anonymous

Those things are not there in our syllabus. We just find the sample space by counting. That's the scope of syllabus till now.

- anonymous

Please help.

- anonymous

Hm fair enough, Good luck :)

- anonymous

bb,gg,bg,gb

- anonymous

dat was answer for 1st ques..

- anonymous

But B1G1, B2G2, that can also come na? I did that way. Salini didi said that.

- anonymous

yeah ...u r right....hwmany marks ques r r these??

- anonymous

itz....bttr if u write ...as u have written

- dumbcow

it doesn't indicate that it matters which boy or which girl is selected
4 possibilities
GG
GB
BG
BB

- anonymous

yeah dumbow is also right...

- dumbcow

and if it did then it would have 20 possibilities
5P2 = 20

- anonymous

I also got 20 but............

- anonymous

c the ques . is 2 children are selected ....and children is classified as male female ...so ans is BB GG BG gb

- anonymous

children are generally clasified as boy girl onle..

- anonymous

rite rite

- anonymous

3 ans ..... 2/8 =1/4

- dumbcow

for #2 does it mean 2 dice are thrown ?

- anonymous

oops srry mate ...i ead it as dice....

- anonymous

thanks...

- dumbcow

i get 4/9 for #2

- anonymous

24/55

- anonymous

?

- anonymous

did u take 0's into account?/

- anonymous

der??

- dumbcow

only numbers divisible be 4or5--> 4,5,8,10,12
3 ways of rolling a 4
4 ways of rolling a 5
5 ways of rolling a 8
3 ways of rolling a 10
1 way of rolling a 12
--> 16/36 = 4/9

- anonymous

i said na ..its not a dice,.

- dumbcow

na??

- anonymous

nope...and second ans is 16/36=4/9

- anonymous

Thanks a lot

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