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[SOLVED] s.aux.26 our-sided die has three blue faces, and one red face......


Well-known member
Jan 31, 2012
four-sided die has three blue face, and one red face.
The die is rolled.
B be the event blue face lies down, and R be the event a red face lands down
a Write down
i $\quad P(B)=\dfrac{3}{4}\quad$ ii $\quad P(R)=\dfrac{1}{4}$

b If the blue face lands down, the dieu is not rolled again. If the red face lands down, the die is rolled once again.
This is represented by the following tree diagram, where p, s, t are probabilities.


Find the value of p, of s and of t.

c Guiseppi plays a game where he rolls the die.
If a blue face lands down, he scores 2 and is finished.
If the red face lands down, he scores 1 and rolls one more time.
Let X be the total score obtained.
$ \quad \texit{
Show that } $P(X=3)=\frac{3}{16}$
[ii] Find $\quad P(X=2)$

[d i] Construct a probability distribution table for X. [5 marks]
[ii] Calculate the expected value of X.

[e] If the total score is 3, Guiseppi wins . If the total score is 2, Guiseppi gets nothing.
Guiseppi plays the game twice. Find the probability that he wins exactly .

ok I only time to do the first question so hope going in right direction
I know the answers to all this is quickly found online but I don't learn too well by C/P
Last edited:

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
a) Yes, the probability of Blue on one roll is 3/4 and the probability if Red is 1/4.

b) On the diagram, p is obviously 3/4. q is (1/4)(3/4)= 3/16. r is (1/4)(1/4)= 1/16.
(Note that 3/4+ 3/16+ 1/16= 1.)

c) The probability of Blue is 3/4 and gives a value 2, The probability of Red, Blue is 3/16 and gives a value 1+ 2= 3. The probability of Red, Red is 1/16 and gives a value 1+ 1= 2. So P(X= 2) is 3/4+ 1/16= 12/16+ 1/16= 13/16. P(X= 3) is 3/16.

di) Since 2 and 3 are the only possible values for X, P(X= 2)= 13/16, P(X= 3)= 3/16 IS the "probability distribution table" for X. (And of course 13/16+ 3/16= 16/16= 1.)

dii) The expected value is (3/4)(2)+ (3/16)(3)+ (1/16)(2)= 24/16+ 9/16+ 2/16= 35/16= 2 and 3/16.

e) The probability Giussepe loses both games is (13/16)(13/16)= 169/256. The probability Giussepe wins one game and losess the other is (13/16)(3/16)+ (3/16)(13/16)= 78/256. The probability Giussepe wins both games is (3/16)(3/16)= 9/256. (Once again, observe that 169/256+ 78/256+ 9/256= 256/256= 1.)
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Well-known member
Jan 31, 2012
that was a great help
ill try the next one all the way thru