toss a fair coin twice and let x be the number of heads obtained.generate the histogram for the distribution.consider tissing the fair coin 3 times,5 times,10 times,and 15 times.generate the histogram for the number of heads also for these cases. as the number of tosses increases what curve can be used to approximate the histogram
Answers
Answer:
ANSWER
(i) When one coin is tosses twice, the sample space is {HH,HT,TH,TT}
Let X represent the number of heads.
∴X(HH)=2,X(HT)=1,X(TH)=1,X(TT)=0
Therefore, X can take the value of 0,1 or 2.
It is known that,
P(HH)−P(HT)−P(TH)−P(TT)=
4
1
P(X=0)=P(TT)= 41
P(X=1)=P(HT)+P(TH)= 41
+ 41
= 21
P(X=2)=P(HH)= 41
Thus, the required probability distribution is as follows.
(ii) When three coins are tossed simultaneously, the sample space is {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
Let X represents the number of tails.
It can be seen that X can take the value of 0,1,2 or 3.
P(X=0)=P(HHH)= 81
P(X=1)=P(HHT)+P(HTH)+P(THH)= 81
+ 81
+ 81
= 83
P(X=2)=P(HTT)+P(THT)+P(TTH)= 81
+ 81
+ 81
= 83
P(X=3)=P(TTT)= 81
Thus, the probability distribution is as follows.
(iii) When a coin is tossed four times, the sample space is
S={HHHH,HHHT,HHTH,HHTT,HTHT,HTHH,HTTH,HTTT,THHH,THHT,THTH,THTT,TTHH,TTHT,TTTH,TTTT}
Let X be the random variable, which represents the number of heads.
It can be seen that X can take the value of 0,1,2,3 or 4.
P(X=0)=P(TTTT)=
16
1
P(X=1)=P(TTTH)+P(TTHT)+P(THTT)+P(HTTT)= 161
+ 161
+ 161
+ 161
= 164
= 41
P(X=2)=P(HHTT)+P(THHT)+P(TTHH)+P(HTTH)+P(HTHT)+P(THTH)= 161
+ 161
+ 161
+ 161
+ 161
+ 161
= 166
= 83
P(X=3)=P(HHHT)+P(HHTH)+P(HTHH)+P(THHH)=161
+ 161
+ 161
+ 161
= 164
=
4
1
P(X=4)=P(HHHH)=
16
1