anonymous
  • anonymous
The probability that a random newborn is male is roughly 0.51 . What is the probability that at least 2 of 4 newborns will be male?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Rizags
  • Rizags
Bernoulli Trials?
Rizags
  • Rizags
Is this for alg 2?
anonymous
  • anonymous
i think is Bernoulli Densities, not alg2 is probability and statistic

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Rizags
  • Rizags
Ok. Either way this will work:
anonymous
  • anonymous
ok
Rizags
  • Rizags
Suppose we have 4 trials, AT LEAST 2 of which are successes. Thus, the probability that we have 2, 3, or 4 successes is 1-P(1)=P(2,3,4). So let's calculate the probability that 1 out of 4 newborns is male. We do that like so:
Rizags
  • Rizags
\[\left(\begin{matrix}4 \\ 1\end{matrix}\right)\times0.51^1\times0.49^3=0.24\]
Rizags
  • Rizags
Finally, to get P(2, 3, or 4), We take 1-P(1):\[\huge 1-P(1)=1-0.24=0.76\]
Rizags
  • Rizags
Make sense?
anonymous
  • anonymous
ya is does make sense, so I need to add 2 3 and 4 ?
Rizags
  • Rizags
Ok, see what we have here is a complement probability. Instead of calculating P(2)+P(3)+P(4), we just calculate P(1) and then subtract it from 1, because:\[\huge \color{blue}{P(2, 3, 4)=1-P(1)}\]
anonymous
  • anonymous
alright
anonymous
  • anonymous
so i do the samething as we didnt with p(1)? 1-p(2), 1-p(3)
Rizags
  • Rizags
no. The answer above is your final answer. There's no Further calculation
anonymous
  • anonymous
also did u subtract .51 to get .49?
Rizags
  • Rizags
P(1)+P(2)+P(3)+P(4)=1
Rizags
  • Rizags
And yes
anonymous
  • anonymous
so 1 2 3 4 need to add up 1 i see
Rizags
  • Rizags
Medal plz
anonymous
  • anonymous
\[P(2) = \left(\begin{matrix}4 \\ 2\end{matrix}\right) * .59^2*.49^2\]
Rizags
  • Rizags
not 0.59 and 0.49, but 0.51 and 0.49
anonymous
  • anonymous
mistype
Rizags
  • Rizags
Remember, for this problem, all you have to do is find P(1), then subtract from 1. THATS IT. If you want to validate then calculate P(2)+P(3)+P(4), which should give you the same answer (0.76)
anonymous
  • anonymous
ok
anonymous
  • anonymous
i got like P(2) 0.37
Rizags
  • Rizags
Yes that's correct. Now do P(3) and P(4) if you want
anonymous
  • anonymous
p(3)=.26 and p(4)= .07
Rizags
  • Rizags
Oh no, I've made a mistake! I LEFT OUT P(0) SORRY!
anonymous
  • anonymous
k
anonymous
  • anonymous
so is 0 1 2 3 and 4
Rizags
  • Rizags
NEW ANSWER CORRECTED:\[\large P(0)+P(1)+P(2)+P(3)+P(4)=1\]
Rizags
  • Rizags
SO\[\large P(2)+P(3)+P(4)=1-(P(0)+P(1))\] AND\[\huge \color {red}{P(0)=0.06}\] AND \[\huge \color{blue}{P(1)=0.24}\]SO\[\large 1-(P(1)+P(2))=1-(0.06+0.24)=0.70\]
Rizags
  • Rizags
That is correct now.
Rizags
  • Rizags
Crap, mistype in response, should say 1-(P(0)+P(1))
anonymous
  • anonymous
ok
anonymous
  • anonymous
somehow i see a equal number .70=.70
anonymous
  • anonymous
found the answer is .702335
Rizags
  • Rizags
yep
anonymous
  • anonymous
.70235

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