Answer to Question #351046 in Statistics and Probability for Bob

Question #351046

2. QUESTION:

A fair coin is tossed, and a fair die is thrown. Write down sample spaces for

(a) the toss of the coin;

(b) the throw of the die;

(c) the combination of these experiments.

Let A be the event that a head is tossed, and B be the event that an odd number is thrown. Directly from the sample space, calculate P(A ∩ B) and P(A ∪ B).


1
Expert's answer
2022-06-27T06:30:50-0400

(a) {HEAD,TAIL}\{HEAD,TAIL\}

(b) {1,2,3,4,5,6}\{1,2,3,4,5,6\}

(c)

{(HEAD,1),(HEAD,2),(HEAD,3),(HEAD,4),(HEAD,5),(HEAD,6),(TAIL,1),(TAIL,2),(TAIL,3),(TAIL,4),(TAIL,5),(TAIL,6)}\{(HEAD,1), (HEAD,2), (HEAD,3),\\ (HEAD,4), (HEAD,5), (HEAD,6),\\ (TAIL,1), (TAIL,2), (TAIL,3),\\ (TAIL,4), (TAIL,5), (TAIL,6)\}

The sample space of combination of the experiments has 1212 elements with equal probabilities 112.\frac{1}{12}.

AB={(HEAD,1),(HEAD,3),(HEAD,5)}A\cap B=\{(HEAD,1), (HEAD,3), (HEAD,5)\},

AB=3|A\cap B|=3, hence P(AB)=3112=14.P(A\cap B)=3\cdot \frac{1}{12}=\frac{1}{4}.

AB={(HEAD,1),(HEAD,2),(HEAD,3),(HEAD,4),(HEAD,5),(HEAD,6),(TAIL,1),(TAIL,3),(TAIL,5)},A\cup B=\{(HEAD,1), (HEAD,2), (HEAD,3),\\ (HEAD,4), (HEAD,5), (HEAD,6),\\ (TAIL,1), (TAIL,3), (TAIL,5)\},

AB=9,|A\cup B|=9, hence P(AB)=9112=34.P(A\cup B)=9\cdot \frac{1}{12}=\frac{3}{4}.

Answer: P(AB)=14,P(AB)=3112=34.P(A\cap B)=\frac{1}{4}, P(A\cup B)=3\cdot \frac{1}{12}=\frac{3}{4}.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment