probability questions (1 Viewer)

[karentao77]

<TABLE class="" style="MARGIN-TOP: 10px" cellSpacing=0 cellPadding=0 width="100%" border=0><TBODY><TR><TD class=table-left rowSpan=2></TD><TD class=bg-icon-question><!-- List Question -->I don't know how to do these, can someone explain please.

1. A bag contains 5 red, 6 blue, 2 white and 7 green balls. If 2 are seleccted at random (without replacement), find the probability of getting at least 1 red ball.

Answer: 17/38

2. In a group of people, 32 are Australian born, 12 were born in Asia, 7 were born in Europe. If 2 of the people are selected at random, find the probability that at least 1 of them will be Australian born.

Answer: 368/425

3. There are 34 men and 32 women at a party. Of these, 13 men and 19 woman are married. If 2 people are chosen at random, find the probability that both will be married.

Answer: 496/2145


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[201055]

Is this in terms of simple probability (ie. P(1R,1 not R) + P (2R)), or using binomial probability (ie. nCr (p)^r (q)^n-r) ?

 

[karentao77]

simple probability

 

[201055]

1) There's a total of 20 balls. Best way to approach this question is to find instead, probability of NOT getting any red balls from both picks,
then 1- P(no red balls on both tries)

1st try: There's 5 red balls possible = (5/20), so (15/20) chance of NOT getting red

2nd try: (without replacement, assuming a NON-RED ball was picked out):
There's 19 balls left.
There's still 5 red balls= 5/19, so (14/19) chance of NOT getting red

Hence, P(not red, not red) = 15/20 * 14/19 = 21/38

So, P (at least one red) = 1 - 21/38 = 17/38

 

[201055]

Same deal with the others:

Q2) Total people = 51 people

P( not aussie born, not aussie born) = 19/51 * 18/50 = 57/425

So, P (at least one aussie born) = 1 - 57/425 = 368/425

Q3) Nice trick question, question tries to confuse you by saying men and women, but the probability asked here is purely: married, or not married

There is a total of 66 people.
There are 32 people married: (13 men + 19 women)
There are 34 people not married.

P( both married) = 32/66 * 31/65 = 496/2145

Hope tat helps

 

[karentao77]

thanksss!!! u rule..!

 

[201055]

P(seed germinating) = 93/100
P (seed not germinating) = 7/100

*This is not a non-replacement probability question, as each seed is independant from each other
= P(only one germinates) = P ( germinate) *P( not germinate)* P (not germinate)
= 93/100 * 7/100 * 7/100
= 0.004557

(Or, 4557/ 1000000)

Edit: lol the question's gone =(

 

[karentao77]

the answer is 1.4%
i got the answer now.. i did 3(0.93 x 0.07 x 0.07)

 

[201055]

Haha well i only saw the question for one second, so was guessing what it was asking, so i think my wrong answer can be excused

 

[karentao77]

hehe, thanks a lot! i kept getting that answer before .. finally realised why.

 

[201055]

Haha probability drove me up the wall last year, really hated it. Bionomial was fun though, the equations made me seem intelligent

 

[karentao77]

i have another probability question.
Most number plates have 3 numbers after 3 numbers. What is the probability of getting ABC 123.
I had this on one of my exams.. and had a mental blank..
for this is it
(1/26)^3 x (1/10)^3 = 1/17576000?
can we use permutations.. and do
26P3 x 10P3 = 1/11232000?
which answer is right? or are they both wrong?

 

[watatank]

the first one looks correct.

 

[karentao77]

argghh.. i did it wrong!!
how do you use permutations to do this?

 

[pixy989]

um well i dont really know.. but err
u can only hav 1 letter or number in each spot right. ABC 123
therefore wouldnt that just mean 26P1 x 26P1 x 26P1 x 10P1 x 10P1 x 10P1?
thats the same as 26^3 x 10^3 i know but, i cant think of any other explanation sorry.



o btw..permutations r part of extension but i can see y u didnt want to put ur questions in separate topics.