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Three different coins are tossed together. Find the probability of getting 

1) exactly two heads 

2) at least 2 heads 

3) at least 2 tails 

 

 

 

 

 
 
 
 
 

Answers (1)

Total outcomes on tossing 3 coins {(HHH),(HHT),(HTT),(THT),(TTH)(HTH),(THH),(TTT)}

Total no. of solutions = n (T) = 8 

1) outcomes with exactly two heads = {(HHT),(HTH),(THH)}

no. of outcomes n (E) = 3 

Probability of getting exactly two heads = P

P = \frac{n (E_1)}{n T} = 3/8 \\\\

2) Outcomes with at least two heads = {(HHH), (HHT),(HTH), (THH)} 

NO. of outcomes = {n (E_2)} = 4

Probability of getting at least two heads = P_2

p_2 = \frac{n ( E _ 2 )}{n T} = 4/8 = 1/2 \\\\

 

3 ) outcomes with at least two tails = {(HTT), (THT),(TTH),(TTT)} 

 

no. of outcomes = n (E_3) = 4 \\\\

Probability of getting at least two tails = \\\\ P_ 3\\\\ P_ 3 = \frac{n (E_3)}{n(T) } = 4/8 = 1/2

Posted by

Ravindra Pindel

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