Expert: Ahmed Salami - 5/4/2010

Question
QUESTION: hi, i need help with this question. its really confusing me.

Three of the six sides of a fair die are green, two are red and one is yellow. If
the die is rolled thrice, what is the probability that all three throws result in the
same colour?

ANSWER: Hi Julie,
Same colour in three throws is GGG or RRR or YYY.
P(G) = 3/6 = 1/2
P(R) = 2/6 = 1/3
P(Y) = 1/6

P(3 same colours) = P(GGG) + P(RRR) + P(YYY)
         = [P(G) x P(G) x P(G)] + [P(R) x P(R) x P(R)] + [P(Y) x P(Y) x P(Y)]
         = [1/2 x 1/2 x 1/2] + [1/3 x 1/3 x 1/3] + [1/6 x 1/6 x 1/6]
         = 1/8 + 1/27 + 1/216
         = 1/6

I've written in short language but you can always get back to me.

Regards

---------- FOLLOW-UP ----------

QUESTION: i also have this question which i am finding hard to do.

It is known that 60% of the Year 12 students of the state of New Sowth Whales(NSW) will enter a university for higher studies after the HSC examination. A random sample of 20 (Year 12) students is to be selected just after their HSC examination. Using only the formula sheet and a calculator, nd the probability that

i. exactly 14 students will enter a university.
ii. at least 18 students will enter a university.
iii. What is the expected number of students entering a university from this sample?

Answer
Hi Julie,
This question involves, and i hope you're familiar with, conditional probability using binomial distribution.
The probability of r successes in n trials with p being the probability of success is given by
P(r) = [nCr][p^r][(1-p)^(n-r)]
where nCr (read as 'n combination r') = n!/(n-r)!r!
Now,
a) p = 60% = 0.6
P(14) = [20C14][0.6^14][0.4^6]
     = (38760)(0.00078)(0.004096)
     = 0.1244
     = 12.44%

b) At least 18 means '18 or 19 or 20'
P(18) = [20C18][0.6^18][0.4^2]
     = (190)(0.000101)(0.16)
     = 0.0031
     
P(19) = [20C19][0.6^19][0.4^1]
     = (20)(0.000061)(0.4)
     = 0.00049
     
P(20) = [20C20][0.6^20][0.4^0]
     = (1)(0.00078)(1)
     = 0.000037
    
P(18 or 19 or 20) = P(18) + P(19) + P(20)
         = 0.0036
         = 0.36%

c) The expected number of students = 60% of 20
         = 0.6 x 20
         = 12

Regards