Medal - 4 fraternal twins! Because there are four babies in the womb, the technician is unable to tell the gender of the babies. (a) Find the probability that the quadruplets will all be boys. Express your answer as a fraction or an exact decimal - do not round. (b) What is the probability that the couple will have 1 boy and 3 girls? (c) What is the probability that the oldest three children will be female?
same as the probability that if you toss 4 coins they all come up head
\[\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\left(\frac{1}{2}\right)^4=\frac{1}{16}\]
is that part a?
yes part A
(b) What is the probability that the couple will have 1 boy and 3 girls? toss four coins what is the probability you get 1 head and 3 tails?
1/3
hmm no
i am assuming you made that number up lets compute it the experiment has \(2^4=16\) possible outcomes, that will be your denominator (not 3) the numerator is the number of ways you can choose one item from a set of 4 sometimes written as \(_4C_1\) or as\(\binom{4}{1}\) or we can list all the possible outcomes if you like
bggg gbgg ggbg gggb
looks good you got 4 of them making your answer \[\frac{4}{16}=\frac{1}{4}\]
so part b is 1/4
yes
what we do for part c
the probability the the first one is a girl is \(\frac{1}{2}\) the probability the the second one is a girl is \(\frac{1}{2}\) the probability the the third one is a girl is \(\frac{1}{2}\) multiply these together to get the probability that all the first three are girls
how do you multiply fractions?
me? i use a calculator
but in general you multiply straight across just like you want to \[\frac{a}{b}\times \frac{c}{d}=\frac{a\times c}{b\times d}\]
1 *1 =1 2*2 = 4
yeah but you have another \(\frac{1}{2}\) there
1/8
\[\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\] right \(\frac{1}{8}\)
so the answer is 1/8?
oui
thank you :))))
yw
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