Question

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?

Answer 1

Bernoulli Trials?

Answer 2

Is this for alg 2?

Answer 3

i think is Bernoulli Densities, not alg2 is probability and statistic

Answer 4

Ok. Either way this will work:

Answer 5

ok

Answer 6

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:

Answer 7

\[\left(\begin{matrix}4 \\ 1\end{matrix}\right)\times0.51^1\times0.49^3=0.24\]

Answer 8

Finally, to get P(2, 3, or 4), We take 1-P(1):\[\huge 1-P(1)=1-0.24=0.76\]

Answer 9

Make sense?

Answer 10

ya is does make sense, so I need to add 2 3 and 4 ?

Answer 11

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:\[\huge \color{blue}{P(2, 3, 4)=1-P(1)}\]

Answer 12

alright

Answer 13

so i do the samething as we didnt with p(1)? 1-p(2), 1-p(3)

Answer 14

no. The answer above is your final answer. There's no Further calculation

Answer 15

also did u subtract .51 to get .49?

Answer 16

P(1)+P(2)+P(3)+P(4)=1

Answer 17

And yes

Answer 18

so 1 2 3 4 need to add up 1 i see

Answer 19

Medal plz

Answer 20

\[P(2) = \left(\begin{matrix}4 \\ 2\end{matrix}\right) * .59^2*.49^2\]

Answer 21

not 0.59 and 0.49, but 0.51 and 0.49

Answer 22

mistype

Answer 23

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)

Answer 24

ok

Answer 25

i got like P(2) 0.37

Answer 26

Yes that's correct. Now do P(3) and P(4) if you want

Answer 27

p(3)=.26 and p(4)= .07

Answer 28

Oh no, I've made a mistake! I LEFT OUT P(0) SORRY!

Answer 29

k

Answer 30

so is 0 1 2 3 and 4

Answer 31

NEW ANSWER CORRECTED:\[\large P(0)+P(1)+P(2)+P(3)+P(4)=1\]

Answer 32

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\[\large 1-(P(1)+P(2))=1-(0.06+0.24)=0.70\]

Answer 33

That is correct now.

Answer 34

Crap, mistype in response, should say 1-(P(0)+P(1))

Answer 35

ok

Answer 36

somehow i see a equal number .70=.70

Answer 37

found the answer is .702335

Answer 38

yep

Answer 39

.70235