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Two events E1 and E2 are called independent if p(E1 â© E2) = p(E1)p(E2). For each of the following pairs of events, which are subsets of the set of all possible outcomes when a coin is tossed three times, determine whether or not they are independent. a) E1: tails comes up with the coin is tossed the first time; E2: heads comes up when the coin is tossed the second time. b) E1: the first coin comes up tails; E2: two, and not three, heads come up in a row. c) E1: the second coin comes up tails; E2: two, and not three, heads come up in a row.

User Hal Eisen
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2 Answers

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Answer:

P (E1) = 18 / 38

P (E2) = 18 / 38

P (E1 and E2) = 10 / 38

Explanation:

User Bbunmp
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7.4k points
6 votes

Answer: a) Independent

b) Independent

c) Dependent

Explanation:

Since, If a coin is tossed three times,

Then, total number of outcomes, n(S) = 8

a)
E_1 : tails comes up with the coin is tossed the first time;


E_1 = { TTT, THH, THT, TTH }


E_2 : heads comes up when the coin is tossed the second time.


E_2 = { THT, HHH, THH, HHT }

Thus,
n(E_1)=4


P(E_1)=(n(E_1))/(n(S))=(4)/(8)=(1)/(2)

Similarly,
P(E_2)=(1)/(2)


P(E_1)* P(E_2)=(1)/(2)* (1)/(2)=(1)/(4)

Since,
E_1\cap E_2 = { THH, THT }


n(E_1\cap E_2) = 2


P(E_1\cap E_2) = (n(E_1\cap E_2))/(n(S))= (2)/(8)=(1)/(4)

Thus,
P(E_1\cap E_2)=P(E_1)*P(E_2)

Therefore,
E_1 and
E_2 are independent events.

B)
E_1 : the first coin comes up tails


E_1 = { TTT, THH, THT, TTH }


E_2 : two, and not three, heads come up in a row


E_2 = { HHT, THH }

Thus,
n(E_1)=4


P(E_1)=(n(E_1))/(n(S))=(4)/(8)=(1)/(2)

Similarly,
P(E_2)=(1)/(4)


P(E_1)* P(E_2)=(1)/(2)* (1)/(4)=(1)/(8)

Since,
E_1\cap E_2 = { THH }


n(E_1\cap E_2) = 1


P(E_1\cap E_2) = (n(E_1\cap E_2))/(n(S))= (1)/(8)

Thus,
P(E_1\cap E_2)=P(E_1)*P(E_2)

Therefore,
E_1 and
E_2 are independent events.

C)
E_1 : the second coin comes up tails;


E_1 = { HTH, HTT, TTT, TTH }


E_2 : two, and not three, heads come up in a row


E_2 = { HHT, THH }

Thus,
n(E_1)=4


P(E_1)=(n(E_1))/(n(S))=(4)/(8)=(1)/(2)

Similarly,
P(E_2)=(1)/(4)


P(E_1)* P(E_2)=(1)/(2)* (1)/(4)=(1)/(8)

Since,
E_1\cap E_2 =
\phi


n(E_1\cap E_2) = 0


P(E_1\cap E_2) = 0

Thus,
P(E_1\cap E_2)\\eq P(E_1)*P(E_2)

Therefore,
E_1 and
E_2 are dependent events.


User Jhogendorn
by
7.7k points
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