Question
Two tetrahedral dice, both numbered 1 to 4 are tossed. If one die us a fair and the other biased so that a four is twice as likely as any other score, setup the possibility space and use it to find the probability that:
a) at least one four is thrown.
b) a total score of four is obtained
So I am confused by the wording in the question, with respect to the word score, the reason being dose the question for a) imply at least a score of 4 is thrown or that one actual 4 is thrown? Because as seen in
b) (which I am not asking to be answered just used for ref) they use the term score in such it mean as combined total i. (1,3) (3,1) ect.
My working for if one biased dice throw a 4.
$P(4)=p$
$P(\neq4)=q$
$p+q=1$
$P(4)=2(P\neq 4)$
$p=2q$
$p=2(1-p)=2-2p$
$p=\frac{2}{3}=P(4)$ ( probability of throwing a 4 on a biased dice)
$P(\neq 4)=\frac{1}{3}$ ( probability of throwing a 4 on a biased dice)
So I calculate the probability of throwing at least 1 is:
$P(x \geq 1)=1-P(none)$
sample space where no 4 is thrown
$(1,1) \ (1,2)\ (1,3)\ (2,1)\ (2,2) \ (2,3) \ (3,1) \ (3,2) \ (3,3)$
So probability of not throwing a 4 at all
$P(none)=\frac{1}{4}\frac{1}{3}+\frac{1}{4}\frac{1}{3}+\frac{1}{4}\frac{1}{3}+\frac{1}{4}\frac{1}{3}+\frac{1}{4}\frac{1}{3}+\frac{1}{4}\frac{1}{3}+\frac{1}{4}\frac{1}{3}+\frac{1}{4}\frac{1}{3}=\frac{3}{4}$
So probability of throwing atleast one 4 is
$P(x \geq)=1-\frac{3}{4}=\frac{1}{4}$
However the ans in the back of the book give 11/20, which make me think I have not visualised the problem correctly.
Best Answer
"One $4$ is thrown" means that one of the dice registers a $4$.
For this the answer is $1/4 + 2/5 - 1/10 = 11/20$. (The probability of getting $4$ on the fair die, plus the probability of getting a $4$ on the unfair die, minus the probability of getting both $4$s.)
"A total score of $4$" means that the two dice add up to $4$.
For this, you have three cases: $1$ is on the fair die, $1$ is on the unfair die, and both $2$.
Hence the probability is $3(1/4)(1/5) = 3/20.$