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Let's now look at the binomial test for the 50% hypothesis for the girls 3 given that boys' heights are distributed normally $\mathcal {n} (68$ inches, $4.5$ inches$)$ and girls are distributed $\mathcal {n} (62$ inches, $3.2$ inches$)$, what is the probability that a girl chosen at random is taller than a boy chosen at random? In fact, 7 girls liked the cake and 1 didn't
Girls (2012)
That's a pretty extreme result for a 50% probability Girls who will attend a party, and features of the party. Is it actually compatible with a true population probability of 50%?
Probability of having 2 girls and probability of having at least one girl ask question asked 8 years, 2 months ago modified 8 years, 2 months ago
Expected girls from one couple$ {}=0.5\cdot1 + 0.25\cdot1 =0.75$ expected boys from one couple$ {}=0.25\cdot1 + 0.25\cdot2 =0.75$ 1 as i said this works for any reasonable rule that could exist in the real world An unreasonable rule would be one in which the expected children per couple was infinite. 1st 2nd boy girl boy seen boy boy boy seen girl boy the net effect is that even if i don't know which one is definitely a boy, the other child can only be a girl or a boy and that is always and only a 1/2 probability (ignoring any biological weighting that girls may represent 51% of births or whatever the reality is). A couple decides to keep having children until they have the same number of boys and girls, and then stop
Assume they never have twins, that the trials are independent with probability 1/2 of a boy, and that they are fertile enough to keep producing children indefinitely. In how many different ways can 5 people sit around a round table Is the symmetry of the table important If the symmetry of the table is not taken into account the.
A couple decides to keep having children until they have at least one boy and at least one girl, and then stop
Assume they never have twi. Suppose i want to build a model to predict some kind of ratio or percentage For example, let's say i want to predict the number of boys vs