In a simultaneous throw of two dice, what is the probability of getting a doublet ?

Answer: Option 4 Getting at most Two heads means 0 to 2 but not more than 2 Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} Let E = event...

Answer: Option 2 In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36 Then, E = {(1, 2), (1, 4), (1, 6), (2, 1), (2,...

Answer: Option 2 In a simultaneous throw of two dice, we have n (S) = (6 × 6) = 36 Let E = event of getting two numbers whose product is...

Answer: Option 1 In a simultaneous throw of two dice, we have n(S) = 6 x 6 = 36 Let E = event of getting two numbers whose sum is odd. Then...

Answer: Option 4 Total 4 cases = [HH, TT, TH, HT] Favourable cases = [HH, TH, HT] Please note we need atmost one tail, not atleast one tail. So probability = $$\frac{{3}}{{4}}$$

Answer: Option 1 In a simultaneous throw of two dice, we have n(s) = 6 × 6 = 36 Let E = event of getting two numbers are same. Then E =...

Answer: Option 4 Total cases = [H, T] - 2 Favourable cases = [H] - 1 So probability of getting head = $$\frac{{1}}{{2}}$$

Answer: Option 4 Total cases = [H, T] - 2 Favourable cases = [T] -1 So probability of getting tails = $$\frac{{1}}{{2}}$$

Answer: Option 3 Sample space (All [possible outcomes)S=(HH, HT, TH, TT) Event(required outcomes) = (TT, HT, TH) $$\eqalign{ & {\text{p}}\left( {\text{E}} \right) = \frac{{{\text{n}}\left( {\text{E}} \right)}}{{{\text{n}}\left( {\text{S}} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{4} \cr}...

Answer: Option 4 Cisco routers, if they do not have the Enterprise edition of the IOS, will default to 5 simultaneous Telnet sessions.