1. ## [SOLVED] Independent Events

hai...stuck ..it simple but i still stuck. hope some one give me guide line how to solve

Question:
A coin is tossed three times. A is the event ' at least two tails' , B isthe event 'three heads or three tails, C is the event 'at least one tail. Which of the following are independent?

(a) A and B
(b) A and C
(c) B and C

solution:
(a) yes
(b) no
(c) no

so how to prove the ans

2. Hi

Compute p(A), p(B), p(C), p(A and B) ...
and show that p(A and B)=p(A).p(B)

3. Originally Posted by nikk
Question:
A coin is tossed three times. A is the event ' at least two tails' , B isthe event 'three heads or three tails, C is the event 'at least one tail. Which of the following are independent?
(a) A and B
(b) A and C
(c) B and C
solution:
(a) yes (b) no (c) no
You need to show that $P(A \cap B)=P(A)P(B)\;\&\;P(A \cap C)
\not{=}P(A)P(C)$

4. My outcomes as below:
HHH,HHT,HTH,HTT,THH,THT,TTH,TTT

Question (a)
i can calculate P(A) = P(HTT) = 1/2 * 1/2 * 1/2 = 1/8
= P(THT) = 1/2 * 1/2 * 1/2 = 1/8
= P(TTH) = 1/2 * 1/2 * 1/2 = 1/8
= P(TTT) = 1/2 * 1/2 * 1/2 = 1/8

There for P(A) = P(HTT) + P(THT) + P(TTH) + P(TTT)
= 1/8 + 1/8 + 1/8 + 1/8
= 4/8
= 1/2
is it correct P(A) = 1/2????

P(B) = P(HHH) + P(TTT)
= 1/8 + 1/8
= 1/4

p(A and B)= P (TTT) + P (TTT)
= 1/8 + 1/8
= 1/4

So, p(A and B) = P(A) * P(B)
1/4 = 1/2 * 1/4
= 1/8

so, not independent but the ans is independent

5. ## logic

is it my equation of P(A) = P(HTT) + P(THT) + P(TTH) + P(TTT)correct

6. Originally Posted by nikk
is it my equation of P(A) = P(HTT) + P(THT) + P(TTH) + P(TTT)correct
Yes it is correct.
But you could make a table, like a truth table.
$\begin{array}{ccccl} H &\vline & H &\vline & H \\ H &\vline & H &\vline & T \\ H &\vline & T &\vline & H \\ H &\vline & T &\vline & T{\color{red}\leftarrow} \\ T &\vline & H &\vline & H \\ T &\vline & H &\vline & T{\color{red}\leftarrow} \\ T &\vline & T &\vline & H{\color{red}\leftarrow} \\ T &\vline & T &\vline & T {\color{red}\leftarrow} \\ \end{array}$

Note that there are $2^3 =8$ rows for the three coins.
Now the counting is easy, see the arrows. We get $\frac{4}{8} = \frac{1}{2}$