Question

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Answer 1

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

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

Let E be the event of getting exactly two heads

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

• Let E be the event of getting exactly two heads

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

We know that, P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

= 3/8

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)

We know that, P(F) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

= 7/8

Answer 2

Answer:

When we toss three coins simultaneously then the possible of outcomes are:

HHH or HHT or HTH or THH or HTT or THT or TTH or TTT respectively;

Where H is denoted for head and T is denoted for tail.

Therefore, total numbers of outcome are

${2}^{3} = 8$

(i) Exactly two heads

Favourable cases ={HHT,HTH,THH}

Probability of getting exactly two heads =

$\frac{3}{8}$

(ii) At least two heads

Favourable cases ={HHH,HHT,HTH,THH}

Probability of getting at least two heads =

$\frac{4}{8} = \frac{1}{2}$

(iii) At least two tails

Favourable cases ={HTT,THT,TTH,TTT}

Probability of getting at least two tails =

$\frac{8}{4} = \frac{2}{1}$

Step-by-step explanation:

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