MHB S.aux.26 our-sided die has three blue faces, and one red face.......

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The discussion revolves around a four-sided die with three blue faces and one red face, focusing on the probabilities associated with rolling the die. The probability of landing on a blue face is 3/4, while the probability of landing on a red face is 1/4. If a blue face lands down, the game ends with a score of 2; if a red face lands down, the player rolls again and can score either 2 or 3. The probability of scoring 3 is calculated as 3/16, while the probability of scoring 2 is 13/16. The expected value of the total score is determined to be 35/16, and the probabilities of winning in multiple games are also analyzed.
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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.

276.png


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
 
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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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Mahalo
that was a great help
ill try the next one all the way thru
 
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