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Answer 1

Answer:

IV) all tails

Step-by-step explanation:

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In tossing three coins, the sample space is given by

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

And, therefore, n(S) = 8.

(i) getting all heads

Let E1 = event of getting all heads. Then,

E1 = {HHH}

and, therefore, n(E1) = 1.

Therefore, P(getting all heads) = P(E1) = n(E1)/n(S) = 1/8.

(ii) getting two heads

Let E2 = event of getting 2 heads. Then,

E2 = {HHT, HTH, THH}

and, therefore, n(E2) = 3.

Therefore, P(getting 2 heads) = P(E2) = n(E2)/n(S) = 3/8.

(iii) getting one head

Let E3 = event of getting 1 head. Then,

E3 = {HTT, THT, TTH} and, therefore,

n(E3) = 3.

Therefore, P(getting 1 head) = P(E3) = n(E3)/n(S) = 3/8.

(iv) getting at least 1 head

Let E4 = event of getting at least 1 head. Then,

E4 = {HTT, THT, TTH, HHT, HTH, THH, HHH}

and, therefore, n(E4) = 7.

Therefore, P(getting at least 1 head) = P(E4) = n(E4)/n(S) = 7/8.

(v) getting at least 2 heads

Let E5 = event of getting at least 2 heads. Then,

E5 = {HHT, HTH, THH, HHH}

and, therefore, n(E5) = 4.

Therefore, P(getting at least 2 heads) = P(E5) = n(E5)/n(S) = 4/8 = 1/2.

(vi) getting atmost 2 heads

Let E6 = event of getting atmost 2 heads. Then,

E6 = {HHT, HTH, HTT, THH, THT, TTH, TTT}

and, therefore, n(E6) = 7.

Therefore, P(getting atmost 2 heads) = P(E6) = n(E6)/n(S) = 7/8

Answer 2

Answer:

When three coins are tossed together, the total number of outcomes =8

i.e., (HHH,HHT,HTH,THH,TTH,THT,HTT,TTT)

Solution (i):

Let E be the event of getting exactly two heads

Therefore, no. of favorable events, n(E)=3(i.e.,HHT,HTH,THH)

$p(e) = \frac{no. \: of \: favoraval \: outcme}{totel \: no. \: of \: outcome} = \frac{3}{8}$

Solution (ii):

Let F be the event of getting atmost two heads

Therefore, no. of favorable events, n(E)=7(i.e.,HHT,HTH,TTT,THH,TTH,THT,HTT)

$p(f) = \frac{7}{8}$

Solution (iii):

Let H be the event of getting at least one head and one tail

Therefore, no. of favorable events, n(H)=6(i.e.,HHT,HTH,THH,TTH,THT,HTT)

$p(h) = \frac{6}{8} = \frac{3}{4}$

Solution (iv):

Let I be the event of getting no tails

Therefore, no. of favorable events, n(I)=1(i.e.,HHH)

$p(h) = \frac{1}{8}$