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zzz [600]
3 years ago
6

Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

the second toss results in heads, and let C be the event that in both tosses the coin lands on the same side. Show that the events A, B, and C are pairwise independent—that is, A and B are independent, A and C are independent, and B and C are independent—but not independent.
Mathematics
1 answer:
aliina [53]3 years ago
6 0

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

A\cap B={HH}

B\cap C={HH}

A\cap C={HH}

P(E)=\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

P(A)=\frac{2}{4}=\frac{1}{2}

Number of favorable cases to event B=2

Number of favorable cases to event C=2

P(B)=\frac{2}{4}=\frac{1}{2}

P(C)=\frac{2}{4}=\frac{1}{2}

If the two events A and B are independent then

P(A)\cdot P(B)=P(A\cap B)

P(A\cap)=\frac{1}{4}

P(B\cap C)=\frac{1}{4}

P(A\cap C)=\frac{1}{4}

P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}

P(B)\cdot P(C)=\frac{1}{4}

P(A)\cdot P(C)=\frac{1}{4}

P(A)\cdot P(B)=P(A\cap B)

Therefore, A and B are independent

P(B)\cdot P(C)=P(B\cap C)

Therefore, B and C are independent

P(A\cap C)=P(A)\cdot P(C)

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

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Find the volume of the solid obtained by rotating the region bounded by y=4x and y=2sqrt(x) about the line x=6.
Pie

Check the picture below.

so by graphing those two, we get that little section in gray as you see there, now, x = 6  is a vertical line, so we'll have to put the equations in y-terms and this is a washer, so we'll use the washer method.

y=4x\implies \cfrac{y}{4}=x\qquad \qquad y=2\sqrt{x}\implies \cfrac{y^2}{4}=x~\hfill \begin{cases} \cfrac{y}{4}=x\\\\ \cfrac{y^2}{4}=x \end{cases}

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