Probability...help! (1 Viewer)

[Liz N]

Can anyone help me with this question below?

A biologist is studying pattern of Male (M) and Female (F) children of families. A family type is designated by a code, for example, FMM denotes a family of 3 children of which the oldest is female and the other two are males. Note that FMM, MFM and MMF are different types. How many family types are there among families with at least one but not more than seven children?

Ans: 254

 

[Liz N]

Oh thankyou!

Now the question looks simple.

 

[Grey Council]

thats what i thought too, but i wasn't sure. ^_^

and probability is like that. Question looks simple after being shown solution. hmph

Anyway, try this one:
From a packet of mixed seed it was estimated that the probability of any seed planted yielding a white carnation was 0.02.
How many seeds must be planted for you to be at least 98% certain of obtaining a white carnation?

lol, first time i saw that, i was like . hehe, its a good question though, i think its the hardest they can ask in 2u maths. So it can certainly be asked in 3u. ^_^

 

[CM_Tutor]

Liz N, this question could easily be extended to the general case (n children, n > 0), where the answer is 2ⁿ⁺¹ - 2.

Grey Council, good question. It is like:

2u HSC, 1988, q10(a) and
2u HSC, 1996, q3(b).

Everyone should have a go at this - you need to be able to solve exponential inequalities correctly.

 

[George W. Bush]

Originally posted by CM_Tutor
Liz N, this question could easily be extended to the general case (n children, n > 0), where the answer is 2ⁿ - 2.

Can you explain? I'm a bit poor at probability.

 

[CM_Tutor]

Firstly, I've edited my post to correct my comment, as the general case is 2ⁿ⁺¹ - 2. Once that is done:

You could sum the GP 2 + 2² + 2³ + ... + 2ⁿ.
We have a = 2, r = 2, and # terns = n. Hence, sum = 2(2ⁿ - 1) / (2 - 1) = 2 * 2ⁿ - 2 * 1 = 2ⁿ⁺¹ - 2.

Alternately, being a bit more rigorous, you could note that, for k children, there could be 0 boys, or 1 boy, or 2 boys, or ..., and hence there are
^kC₀ + ^kC₁ + ^kC₂ + ... + ^kC_k possibilities. This sums to 2^k, by expanding (1 + x)^k and taking x = 1. This allows you to generate the GP, and then proceed as before.

 

[Liz N]

CM_Tutor, I did your method too, but it turn out to be complicated and errors. When I did again, I realise I done a silly mistake.

I hate silly mistakes!

Grey council, whats the answer for that question? I did it, but wasn't sure if my answer was right.

 

[George W. Bush]

194?

 

[Xayma]

Originally posted by Grey Council
From a packet of mixed seed it was estimated that the probability of any seed planted yielding a white carnation was 0.02.
How many seeds must be planted for you to be at least 98% certain of obtaining a white carnation?

Let the number of seeds be n.

Now the probability of no white carnation after n seeds is 2% or 0.02
Now the probability of no white carnation after 1 seed is 0.98

Therefore 0.02=0.98ⁿ
log 0.02=log 0.98ⁿ
log 0.02=n log 0.98
n=log 0.02/log0.98
n=194 (rounding up because it is at least)

 

[CM_Tutor]

Xayma has the correct answer (as did GWB), but the question really should be done using an inequality, and so a better answer (IMO) would be:

From n seeds, P(at least 1 white carnation) = 1 - P(no white carnations) = 1 - 0.98ⁿ

We want P(at least 1 white carnation) > 98 %
So, 1 - 0.98ⁿ > 0.98
0.02 > 0.98ⁿ
ln 0.02 > ln 0.98ⁿ
ln 0.02 > n * ln 0.98
ln 0.02 / ln 0.98 < n, as ln 0.98 < 0
n > 193.63 ...

So, we must plant 194 (or more) seeds to be at least 98 % certain of producing at least 1 white carnation.

 

[Grey Council]

sorry, was away during weekend.

But yes, that is correct. Obviously, CM_Tutor did it.

and I like that question too. Glad you liked it.

 

[Liz N]

Heres another question...

Five of the 7 letters of the word CRICKET are selected and arranged in a row. How many different arrangements are possibe?

Im not sure if the answer is 1320 coz there is no solution.

 

[George W. Bush]

 

[CM_Tutor]

Liz N, I agree with you - I think it is 1320 because:

there are 5! words that can be made using the letters R, I, K, E and T

there are ⁵C₄ * 5! words that can be made using a C and four of the letters R, I, K, E and T

there are ⁵C₃ * 5! / 2! words that can be made using both C's and three of the letters R, I, K, E and T

So, total number of words = 5! + ⁵C₄ * 5! + ⁵C₃ * 5! / 2! = 1320

 

[Grey Council]

nice question that.

Its question 7/8 from a 4u paper though. lol

 

[Liz N]

Im confident with the answer 1320. Thanx.. for all your help.

 

[-=«MÄLÅÇhïtÊ»=-]

ye its 1320

 

[Liz N]

Guys can u plz help me with this question:

Find the number of possible arrangements and selections of 4 letters of the word PROPORTION.

Ans: arrangements-758, selections- 53

 

[CM_Tutor]

# selections

There are ⁶C₄ = 15 selections with 4 different letters, as there are 6 letters to choose from (P, R, O, T, I and N).

There are ³C₁ * ⁵C₂ = 30 selections with 3 different letters, as 1 of 3 possible letters (P, R, O) will appear twice, and two other letters are chosen from T, I, N, and whichever two from P, R and O were not chosen as the double letter.

If there are only two different letters, then there are ³C₂ = 3 selections possible with 2 letters repeated twice (ie PPRR, for example), and ⁵C₁ = 5 selections with one letter appearing 3 times.

Thus, the total number of selections possible = 15 + 30 + 3 + 5 = 53

# arrangements

To count arrangements, we must look at all arrangements possible for each selection. I get this to be:

⁶C₄ * 4! + ³C₁ * ⁵C₂ * 4! / 2! + ³C₁ * 4! / (2!)² + ⁵C₁ * 4! / 3! = 360 + 360 + 18 + 20 = 758, as expected.

 

[Liz N]

Im stuck on this question... can anyone help?

If over a certain period of the year, rain falls at random and an average of 12 days in every 30, find the probability that
a) the first three days on a given week will be fine and the remainder wet
b) rain will fall on just 3 days of a given week,
c) at least 3 days in a given week will be fine?

Ans: 0.0055, 0.2903, 0.9037