Question

challenge! ready?

Answer 1

solve without using complement, that is \(\sf P(x) = 1 - P(x') \)

Answer 2

1/5?

Answer 3

@UsukiDoll

Answer 4

Probability of at least 1 girl
\[P(X \ge 1)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\]

Answer 5

I don't do stats problems.

Answer 6

probability of 1 girl + probability of 2 girls + probability of 3 girls + probability of 4 girls + probability of 5 girls

Answer 7

continue

Answer 8

can i do it as well

Answer 9

are you using \(\large \cup_{x=1} ^{n} X_i = P(x_1 \cup x_2 \cup x_3\cup x_4 \cup x_5) \)

Answer 10

\[P(X \ge 1)=\frac{1}{32}+\frac{1}{16}+\frac{1}{8}+\frac{1}{4}+\frac{1}{2}=\frac{1+2+4+8+16}{32}=\frac{31}{32}\]
another method
\[P(X=0)=\frac{1}{32}\]\[P(X \ge 1)=1-P(X=0)=1-\frac{1}{32}=\frac{32-1}{32}=\frac{31}{32}\]

Answer 11

haha wow

Answer 12

i said without using complement

Answer 13

I've done both with and without compliment, the compliment method is much faster anyway

Answer 14

another equivalent reasoning, can be this:
the even space contains 32 possible events, nevertheless only one event, of them, is like this:
BBBBB,
where B stands for boy, so the total number of favorable events is 31

Answer 15

I think you can also do this using binomial distribution
where n=5 p=q=0.5

Answer 16

why do u people make ur life so hard

Answer 17

oops..event space