We received a large number of responses of
excellent quality.
Ben from St Peter's followed the tree
diagram and calculated out the answer:
If you flip a coin three times the chance of getting at least one
head is 87.5%. To get this outcome I used the provided tree
diagram to establish how many outcomes used one head.
Llewellyn from St Peter's and Diamor from
Willington County Grammar School both observed an interesting
pattern and expanded the answer to flipping ten coins:
If you flip a coin 3 times the probability of getting at least one
heads is 7 in 8 by reading the table. This table also works the
opposite way, the chances of Charlie getting no heads is 1 in 8
because out of all the outcomes only one of them has only tails. I
notice that if you add these probabilities together you get the
total amount of outcomes (7+1=8). If you flip a coin 4 times the
probability of you getting at least one heads is 15 in 16 because
you times the amount of outcomes you can get by flipping 3 coins by
2, it results in 16 and then you minus 1 from it. With 5 coins to
flip you just times 16 by 2 and then minus 1, so it would result
with a 31 in 32 chance of getting at least one heads. With 6 coins
you times by 2 and minus by 1 again resulting in a 63 in 64 chance.
To find the chance of getting at least one heads if you flip ten
coins you times 64 by 2 four times or by 16 once and then minus 1,
this results in a 1063 in 1064 chance of getting at least one
heads.
Neeraj from Wilson School developed a
generalization for different numbers of possible
outcomes:
I noticed that when you add the probabilities together they make a
whole. A quick way of figuring out how many times you get at least
one head is, that it is always the no. Of possible outcomes minus
one over the no. of possible outcomes So: if No of possible
outcomes = n the equation would be: P= (n- 1)/n
One student suggested how to calculate the
number of desired outcomes:
If the number of times flipped =p Then the number of outcomes that
contain a head is$2^p-1$
So for flipping a coin $10$ times, the number of outcomes with at
least one head is $2^{10}-1 = 1024 - 1 = 1023$
Luke from Maidstone Grammar School
went further to investigate the next part of the
question:
When there are 4 green balls in the bag and there are 6 red balls
the probability of randomly selecting a green ball is 0.4
($\frac{2}{5}$) and the probability selecting a red ball is 0.6
($\frac{3}{5}$).
If a ball is selected and then
replaced the probability of picking a red ball or a green
ball is the same every time. When 3 balls are picked with
replacement the probability of getting at least one green is
1-(the probability of getting 3 reds)
Because the probability is the same every time the chance of
getting 3 reds is $0.6^3=0.216$ (or in fractions $(\frac{3}{5})^3 =
\frac{27}{125}$). So the probability of getting at least one green
is $1-0.216=0.784$ (or in fractions $1 - \frac{27}{125} =
\frac{98}{125}$).
When the balls are not
replaced the probability of getting at least one green is
still 1-(the probability of getting 3 reds). In each draw the
probability of drawing a red ball is $\frac{\text{the number of red
balls}}{\text{the total number of balls}}$
On the first draw there are 6 red balls out of 10 so the
probability of picking a red is $\frac{6}{10}$.
On the second draw there are 5 red balls out of 9 so the
probability of picking a red is $\frac{5}{9}$.
On the final draw there are 4 red balls out of 8 so the probability
of picking a red is $\frac{4}{8}$.
The probability of this sequence of draws happening is the
probability of each draw multiplied together. i.e.:
$\frac{6}{10}\times\frac{5}{9}\times\frac{4}{8}=\frac{1}{6}$
The probability of drawing all reds is $\frac{1}{6}$ and so the
probability of drawing at least one green is $\frac{5}{6}$.
Helen from Stroud finished up the
problem:
When children are selected for the school council they are not
replaced. The children are selected one after another and each time
the probability of a boy being selected is
P(boy selected first) = $\frac{\text{the number of boys in the
class}}{\text{the total number of children in the class}}$
Note: the class refers to students who have not already been made
part of the council.
To find the probability that there will be at least one boy, find
the probability that all three are girls, and then
P(at least one boy selected) = 1-P(all girls selected)
to get the answer.
The probability of picking a girl is
P(girl selected first) = $\frac{\text{number of girls in
class}}{\text{total number in class}}= \frac{15}{28}$
Then P(second selected also a girl) = $\frac{14}{27}$
And P(third selected also a girl) = $\frac{13}{26}$
So P(all girls selected) =
$\frac{15}{28}\times\frac{14}{27}\times\frac{13}{26} =
\frac{5}{36}$
Then the answer is
P(at least one boy selected) = 1 - P(all girls selected) = 1 -
$\frac{5}{36}$ = $\frac{31}{36}$