A coin is tossed three times.
Event A: two heads appear
Event B: last should be head
Then identify whether events A and B are independent or not.
The sample space of the experiment is `S = {HHH,HHT,HTH,THH,HT T,THT,T TH,T T T}` `A = {T T T}` `B = {HT T, THT, T TH}` `C = {HHT, HTH,...
2 Answers 1 viewsIf a coin is tossed three times, then the sample S is `S={HHH,HHT, HTH, HT T ,THH,THT, T TH, T T T }` `thereforen(S)=8.` (i) `A={HHH, HTH, THH, T TH]`...
2 Answers 1 views`P(E_(1))=1-["P(all heads)+P(all tails")]` `=1-[(1)/(2^(n))+(1)/(2^(n))]=1-(1)/(2^(n-1))` `P(E_(2))=P("no head)+P(exactly one head")` `=(1)/(2^(n))+^(""n)C_(1).(1)/(2^(n))=(n)/(2^(n))` `P(E_(1)nnE_(2))=("exactly one head and"(n-1)till)` `=^(""n)C_(1).(1)/(2).(1)/(2^(n-1))=(n)/(2^(n))` If `E_(1)and E_(2)` are indipendent, then `P(E_(1)nnE_(2))=P(E_(1)).P(E_(2))` `implies(n)/(2^(n))=(1-(1)/(2^(n-1)))((n+1)/(2^(n)))` `impliesn=(1-(1)/(2^(n-1)))(n+1)`
2 Answers 1 viewsCorrect Answer - `3//4` `P(E)=4/8=1/2{because"favorable cases are"THH, HTH, HHT, HHH]` `P(F)=4/8=1/2[because"favorable cases are"HTH, HHT,HT T, HHH]` `P(EnnF)=3/8` `P(E//F)=(P(EnnF))/(P( F))=(3//8)/(1//2)=3/4`
2 Answers 1 viewsCorrect Answer - `n=14` According to question `""^(n)C_(6)((1)/(2))^(6)((1)/(2))^(n-6)=""^(n)C_(8)((1)/(2))^(8)((1)/(2))^(n-8)` `or ""^(n)C_(6)((1)/(2))^(n)=""^(n)C_(8)((1)/(2))^(n)` `or""^(n)C_(6)=""^(n)C_(8)=""^(n)C_(n-8)` `implies6=n-8orn=14`
2 Answers 1 viewsCorrect Answer - D Let `p_(i)` denote the probability that out of 10 tosses, head occurs I times and no two heads occur consecutively. If is clear that `igt5.` For I...
2 Answers 1 viewsCorrect Answer - C The required probability is 1- probability of getting equal number of heads and tails `=1-""^(2n)C_(n)((1)/(2))^(n)((1)/(2))^(2n-n)` `=1-((2n)!)/((n!)^(2))xx(1)/(4^(n))`
2 Answers 1 viewsCorrect Answer - 2 Here `P(E)=1/2and P(F_(k))=""^(n)C_(k).(1)/(2^(n))` Also, `P(EnnF_(k))=p` (exactly k heads are obtained and head obtained in first filp) `=1/2""^(n-1)C_(k-1)((1)/(2))^(n-1)` Events E and `f_(k)` are independent. Therefore, `P(EnnF_(k))=P(E).P(F_(k))` `or ""^(n-1)C_(k-1)xx(1)/(2^(n))=1/2xx""^(n)C_(k)(1)/(2^(n))`...
2 Answers 3 viewsCorrect Answer - 3 `P(E_(1))=1-[P ("all heads)=P(all tails")]` `=1-[(1)/(2^(n))+(1)/(2^(n))]=1-(1)/(2^(n-1))` `P(E_(2))=["P(no head)+P (exactly one head)"]` `=(1)/(2^(n))+""^(n)C_(1)xx(1)/(2^(n-1))=(n)/(2^(n))` If `E_(1) and E_(2)` are independent, then `(n)/(2^(n))(1-(1)/(2^(n-1)))((n+1)/(2^(n)))` `orn =(1-(1)/(2^(n-1)))(n+1)` `or n=n+1-(n+1)/(2^(n-1))` `or n+1=2^(n-1)` `or n=3`
2 Answers 1 viewsCorrect Answer - `(63)/(100)`
2 Answers 1 views