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
B. is reasonable because 9.2×4.8 = 44.16
What is the question? i could help you if i knew the question
1m is equal to 100cm, so the scale reduces the size of the actual object
1st year 500* 0.03 = 15+500=515
2nd year 515*0.03= 15.45 + 515 = 530.45
3rd year 530.45*0.03= 15.9135 + 530.45 = 546.3635
4th year 546.3635 * 0.03 = 16.390905 + 546.3635 = 562.754405
562.754405
