Answer:
25%
Explanation:
the probability at least two of them will have type O can be solved as follows
Let the chances be
(Chance of Alex is I^B I ) x (1-none have type O) X (Chance of Anna is I^A i)
=>>(2/3) x (1-.75 3 ) x (2/3)= 0.25
So 0.25x 100 = 25%
The blood type of the offspring is dependent on the probable blood type
their parents.
The probability that at least 2 of them will have type O is [tex]\underline{\frac{5}{144}}[/tex]
Reasons:
Alex; Both parents are [tex]I^Bi[/tex]
Anna; Both parents are [tex]I^Ai[/tex]
Alex
[tex]\left[\begin{array}{lll}&I^B&i\\I^B&I^BI^B&I^Bi\\i&I^Bi&ii\end{array}\right][/tex]
Anna
[tex]\left[\begin{array}{lll}&I^A&i\\I^A&I^AI^A&I^Ai\\i&I^Ai&ii\end{array}\right][/tex]
Probability that [tex]I^Bi[/tex] from Alex = [tex]\frac{2}{3}[/tex]
Probability that [tex]I^Ai[/tex] from Anna = [tex]\frac{2}{3}[/tex]
Probability that two have type O = [tex]\frac{2}{3}[/tex] × [tex]\frac{2}{3}[/tex] × [tex]\left(\frac{1}{4}\right)^2[/tex] = [tex]\frac{1}{36}[/tex]
Probability that three have type O = [tex]\frac{2}{3}[/tex] × [tex]\frac{2}{3}[/tex] × [tex]\left(\frac{1}{4}\right)^3[/tex] = [tex]\frac{1}{144}[/tex]
Probability that at least two have type O is
[tex]P(At \ least \ 2)=\dfrac{1}{144} + \dfrac{1}{36} = \dfrac{5}{144}[/tex]
The probability that at least 2 of them will have type O is [tex]\underline{\frac{5}{144}}[/tex]
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