Try this option:
1] P(S)=0.84;
2] P(S or C, but not both)=0.4;
3] P(C)=0.76;
4] P(S∪C)=0.6;
5] P(C, but not S)=0.16;
6] P(S∩C)=1.00.
Answer:
Our answer is 0.8172
Step-by-step explanation:
P(doubles on a single roll of pair of dice) =(6/36) =1/6
therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)
=1-(1-1/6)3 =91/216
for 12 players this follows binomial distribution with parameter n=12 and p=91/216
probability that at least 4 of the players will get “doubles” at least once =P(X>=4)
=1-(P(X<=3)
=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)
=1-0.1828
=0.8172
Answer:
Step-by-step explanation:
-(11,14,25) is the correct answer. the two shorter sides must add to a longer distance than the longest side.
