Question

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.

Answer 1

Answer:  a) Independent

b) Independent

c) Dependent

Step-by-step 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, $n(E_1)=4$

⇒  $P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

Similarly,  $P(E_2)=\frac{1}{2}$

⇒  $P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

Since,   $E_1\cap E_2$ = { THH, THT }

$n(E_1\cap E_2) = 2$

⇒ $P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{2}{8}=\frac{1}{4}$

Thus,  $P(E_1\cap E_2)=P(E_1)\timesP(E_2)$

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, $n(E_1)=4$

⇒  $P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

Similarly,  $P(E_2)=\frac{1}{4}$

⇒  $P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$

Since,   $E_1\cap E_2$ = { THH }

$n(E_1\cap E_2) = 1$

⇒ $P(E_1\cap E_2) = \frac{n(E_1\cap E_2)}{n(S)}= \frac{1}{8}$

Thus,  $P(E_1\cap E_2)=P(E_1)\timesP(E_2)$

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, $n(E_1)=4$

⇒  $P(E_1)=\frac{n(E_1)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

Similarly,  $P(E_2)=\frac{1}{4}$

⇒  $P(E_1)\times P(E_2)=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$

Since,   $E_1\cap E_2$ = $\phi$

$n(E_1\cap E_2) = 0$

⇒ $P(E_1\cap E_2) = 0$

Thus,  $P(E_1\cap E_2)\neq P(E_1)\timesP(E_2)$

Therefore,  $E_1$ and $E_2$ are dependent events.