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Brut [27]
2 years ago
15

Assuming boys and girls are equally​ likely, find the probability of a couple having a baby girl when their sixth child is​ born

, given that the first five children were all girls.
Mathematics
1 answer:
Ivanshal [37]2 years ago
3 0

Answer:  The required probability of having 6th girl is 0.5.

Step-by-step explanation:  Given that boys and girls are equally likely.

We are to find the probability of a couple having a baby girl when their sixth child is​ born, given that the first five children were all girls.

Since the events of having a boy and a girl are independent of each other, so

the probability of having 6th girl dose not depend on the birth of the first five girls.

We know that there are only two possible cases (either a boy or girl will born).

So, sample space, S = {G, B}  and the event E of having a girl is, E = {G}.

That is, n(S) = 2 and n(E) = 1.

Therefore, the probability of event E is given by

P(E)=\dfrac{n(E)}{n(S)}=\dfrac{1}{2}=0.5.

Thus, the required probability of having 6th girl is 0.5.

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Use geometric series to find the fraction of 0.8967898989
Sliva [168]

I'm guessing the repeating part is 89 at the end, so that

x=0.8967\overline89\implies10^4x=8967.\overline{89}

Then

10^4x=8967+\displaystyle89\sum_{i=1}^\infty\frac1{100^i}

10^4x=8967+89\left(\dfrac1{1-\frac1{100}}-1\right)

10^4x=8967+\dfrac{89}{99}

x=\dfrac{8967}{10^4}+\dfrac{89}{99\cdot10^4}

x=\dfrac{443911}{495000}

###

An arguably quicker way without using geometric series:

10^4x=8967.\overline{89}

10^6x=896789.\overline{89}

10^6x-10^4x=887822

x=\dfrac{887822}{10^6-10^4}=\dfrac{443911}{495000}

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