Answer:
b nhbhbjhjbhjbhbjbhjh
Step-by-step explanation:
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
The answer is
<span>x+1</span>, <span>x+2</span>, <span>x+3</span>, <span>x+4</span> and <span>x+5</span>.
The sum of these six integers is 393 so we can write:
<span>x+x+1+x+2+x+3+x+4+x+5=393</span>
<span>6x+1+2+3+4+5=393</span>
<span>6x+15=393</span>
<span>6x+15−15=393−15</span>
<span>6x+0=378</span>
<span><span><span>6x</span>6</span>=<span>3786</span></span>
<span>x=63</span>
Because the first integer is 63 then the third would be <span>x+2</span> or <span>63+2=<span>65</span></span>
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<span><span>Hope its helps :)</span></span>
Not sure but the answer may be M=4
B=2
