# Thread: Binomial Distribution?

1. ## Binomial Distribution?

Hi, should I use binomial distribution to solve the following question? (a) to be specific. I managed to solve (b)

Research shows that 40% of Singaporeans prefer tea, 35% prefer coffee and the rest prefer milk. 9 persons were randomly chosen. Find the probability that

(a) the same number of people prefer each type of drink. Ans:0.07203

(b) more people prefer tea to the other drinks. Ans:0.2666

T~B(9,0.4)
C~B(9,0.35)
M~B(9,0.25)

P(T=3) x P(C=3) x P(M=3)
= {9C3(0.4)^3 x (0.6)^6} x {9C3(0.35)^3 x (0.65)^6} x {9C3(0.25)^3 x (0.75)^6}
= 0.01591

But the answer given is 0.07203. Spot any mistakes?
Thanks for your help.

2. Originally Posted by cyt91
Hi, should I use binomial distribution to solve the following question? (a) to be specific. I managed to solve (b)

Research shows that 40% of Singaporeans prefer tea, 35% prefer coffee and the rest prefer milk. 9 persons were randomly chosen. Find the probability that

(a) the same number of people prefer each type of drink. Ans:0.07203

(b) more people prefer tea to the other drinks. Ans:0.2666

T~B(9,0.4)
C~B(9,0.35)
M~B(9,0.25)

P(T=3) x P(C=3) x P(M=3)
= {9C3(0.4)^3 x (0.6)^6} x {9C3(0.35)^3 x (0.65)^6} x {9C3(0.25)^3 x (0.75)^6}
= 0.01591

But the answer given is 0.07203. Spot any mistakes?
Thanks for your help.
Multinomial distribution - Wikipedia, the free encyclopedia

3. You can get there using the Binomial Distribution.

The probability that precisely 3 people prefer tea is $\displaystyle 9C3(0.4)^{3}(0.6)^{6}.$

The probabilty that of the remaining 6, precisely 3 prefer coffee is $\displaystyle 6C3(35/60)^{3}(25/60)^{3}.$

Since you need both of these to happen you simply multiply them. You don't have to worry about the 'milkies' they will just be the remaining 3.

4. Thanks for your reply. Why can't I reach the same answer with my (wrong) method?
What is the problem with my approach?

5. Basically, you are counting the same people twice or three times.
Your line P(T=3)*P(C=3)*P(M=3) should be read as, the probability that exactly three people drink tea AND THEN (of the remainder), exactly three people drink coffee AND THEN (of the remainder) exactly three people drink milk.
The final probability P(M=3) will be equal to 1 because the tea and coffee drinkers have been removed, the three that are left are bound to prefer milk. The probability for the coffee drinkers is of a three from six situation.
You should be able to arrive at the same result by carrying out the calculation in a different order. That is, working out say (using your notation) P(M=3)*P(C=3)*P(T=3) or P(C=3)*P(M=3)*P(T=3) etc (six possible different orders) should always give you the same result, and in each case the third of the probabilities will be 1.

6. Thanks a lot. You've been very helpful.

7. Hello, cyt91!

Should I use binomial distribution to solve the following question?

With three events, we must use a multinomial distribution.

Research shows that 40% prefer tea, 35% prefer coffee and the rest prefer milk.
Nine persons were randomly chosen. .Find the probability that

(a) the same number of people prefer each type of drink. .Ans: 0.07203

$\displaystyle \displaystyle P(\text{3 of each}) \;=\;{9\choose3,3,3}(0.40)^3(0.35)^3(0.25) \;=\;0.07203$

Your answer should have been derived something like this . . .

We want three of the nine people to drink Tea: .$\displaystyle \displaystyle {9\choose3}(0.4)^3$

Of the remaining six people,
. . we want three Coffee and three Milk: .$\displaystyle \displaystyle {6\choose3}(0.35)^3(0.25)^3$

The answer is: .$\displaystyle \displaystyle {9\choose3}(0.6)^3{6\choose3}(0.35)^3(0.25)^3 \;=\;0.07203$

(b) more people prefer tea to the other drinks. . Ans:0.2666

$\displaystyle P(\text{Tea}) \:=\:0.4,\;\;P(\text{Other}) \:=\:0.6$

$\displaystyle \begin{array}{ccccc} P(\text{9 Tea, 0 Other}) &=& {9\choose9}(0.4)^9(0.6)^0 &=& 0.000\,262\,144 \\ \\ [-3mm] P(\text{8 Tea, 1 Other}) &=& {9\choose8}(0.4)^8(0.6)^1 &=& 0.003\,538\,944 \\ \\ [-3mm] P(\text{7 Tea, 2 Others}) &=& {9\choose7}(0.4)^7(0.6)^2 &=& 0.021\,233\,664 \\ \\ [-3mm] P(\text{6 Tea, 3 Others}) &=& {9\choose6}(0.4)^6(0.6)^3 &=& 0.074\,317\,824 \\ \\ [-3mm] P(\text{5 Tea, 4 Others}) &=& {9\choose5}(0.4)^5(0.6)^4 &=& 0.167\,215\,104 \\ \\[-4mm] \hline \\[-4mm] && \text{Total:} && 0.266\,567\,680\end{array}$

Therefore: .$\displaystyle P(\text{more Tea}) \:\approx\:0.2666$

8. Thanks a lot. You guys have been extremely helpful. Keep up the good work!