Question #88349
A fair coin is tossed five times. Find the possibilities that a head appears
i) exactly three times
ii) at least two times
iii) at the most four times.
1
Expert's answer
2019-04-29T12:01:57-0400

Solution. Let event A be the appearance of a head. The probability of event A is 0.5 for each throw. Since the probability is the same for all throws, we use the formula


Pn(m)=CnmpmqnmP_n(m) = C_n^m*p^m*q^{n-m}

where n=5 is number of events; m is number of event A; C is number of combinations; p=0.5 is probability of the event A; q=1-p=0.5. Find the probability for all possible quantities of event A with 5 throws. Get


P5(0)=C500.500.55=0.03125P_5(0) = C_5^0*0.5^0*0.5^5=0.03125

P5(1)=C510.510.54=50.03125=0.15625P_5(1) = C_5^1*0.5^1*0.5^4=5*0.03125=0.15625

P5(2)=C520.520.53=100.03125=0.31250P_5(2) = C_5^2*0.5^2*0.5^3=10*0.03125=0.31250


P5(3)=C530.530.52=100.03125=0.31250P_5(3) = C_5^3*0.5^3*0.5^2=10*0.03125=0.31250


P5(4)=C540.540.51=50.03125=0.15625P_5(4) = C_5^4*0.5^4*0.5^1=5*0.03125=0.15625

P5(5)=C550.550.50=0.03125P_5(5) = C_5^5*0.5^5*0.5^0=0.03125

Therefore

i)


P5(3)=0.31250P_5(3) =0.31250

ii)

P5(2)+P5(3)+P5(4)+P5(5)=0.8125P_5(2)+P_5(3) +P_5(4)+P_5(5)= 0.8125

iii)

P5(0)+P5(1)+P5(2)+P5(3)+P5(4)=0.96875P_5(0)+P_5(1) +P_5(2)+P_5(3)+P_5(4)=0.96875

Answer.  i) 0.3125 ii) 0.8125 iii) 0.96875




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