We received a big number the responses ofexcellent quality. Ben from St Peter"s adhered to the treediagram and also calculated out the answer:If you flip a coin 3 times the chance of obtaining at the very least onehead is 87.5%. To gain this outcome I used the provided treediagram to establish how plenty of outcomes used one head.Llewellyn native St Peter"s and also Diamor fromWillington ar Grammar institution both observed an interestingpattern and expanded the answer to flipping ten coins:If you flip a coin 3 time the probability of getting at the very least oneheads is 7 in 8 by analysis the table. This table also works theopposite way, the chances of Charlie gaining no heads is 1 in 8because the end of every the outcomes only among them has actually only tails. Inotice that if you add these probabilities together you obtain thetotal amount of outcomes (7+1=8). If you flip a coin 4 times theprobability of you obtaining at least one heads is 15 in 16 becauseyou times the lot of outcomes friend can obtain by flipping 3 coins by2, it outcomes in 16 and also then girlfriend minus 1 indigenous it. Through 5 coins toflip you just times 16 by 2 and then minus 1, for this reason it would resultwith a 31 in 32 opportunity of obtaining at least one heads. V 6 coinsyou times by 2 and minus by 1 again resulting in a 63 in 64 chance.To find the opportunity of gaining at the very least one top if you upper and lower reversal tencoins you times 64 by 2 four times or by 16 once and also then minus 1,this outcomes in a 1063 in 1064 opportunity of obtaining at least oneheads.  Neeraj indigenous Wilson School developed ageneralization for different numbers that possibleoutcomes: i noticed that once you include the probabilities with each other they do awhole. A quick method of figuring out how numerous times you obtain at leastone head is, the it is constantly the no. Of feasible outcomes minusone end the no. Of feasible outcomes So: if No of possibleoutcomes = n the equation would be: P= (n- 1)/n  One student suggested how to calculation thenumber of preferred outcomes:If the variety of times flipped =p climate the variety of outcomes thatcontain a head is$2^p-1$So because that flipping a coin $10$ times, the variety of outcomes through atleast one head is $2^10-1 = 1024 - 1 = 1023$ Luke from Maidstone Grammar Schoolwent more to inspection the next part of thequestion:When there room 4 green balls in the bag and there are 6 red ballsthe probability of randomly selecting a green ball is 0.4($\frac25$) and the probability choosing a red sphere is 0.6($\frac35$).If a sphere is selected and also thenreplaced the probability of picking a red round or a greenball is the exact same every time. Once 3 balls space picked withreplacement the probability of getting at least one eco-friendly is1-(the probability of getting 3 reds)Because the probability is the exact same every time the possibility ofgetting 3 reds is $0.6^3=0.216$ (or in fractions $(\frac35)^3 =\frac27125$). So the probability of getting at the very least one greenis $1-0.216=0.784$ (or in fractions $1 - \frac27125 =\frac98125$). When the balls are notreplaced the probability of getting at least one eco-friendly isstill 1-(the probability of getting 3 reds). In each draw theprobability of illustration a red sphere is $\frac\textthe number of redballs\textthe total number of balls$On the very first draw there are 6 red balls the end of 10 for this reason theprobability of choose a red is $\frac610$.On the second draw there space 5 red balls the end of 9 therefore theprobability of choose a red is $\frac59$.On the final attract there space 4 red balls the end of 8 for this reason the probabilityof choose a red is $\frac48$.The probability that this succession of draws happening is theprobability that each attract multiplied together. I.e.:$\frac610\times\frac59\times\frac48=\frac16$The probability of illustration all reds is $\frac16$ and also so theprobability of drawing at least one eco-friendly is $\frac56$. Helen native Stroud perfect up theproblem: When kids are selected for the school council they are notreplaced.

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The kids are selected one after ~ another and each timethe probability that a boy being selected isP(boy selected first) = $\frac\textthe number of boys in theclass\textthe total number of children in the class$Note: the class refers to students who have not currently been madepart of the council.To find the probability the there will be at the very least one boy, findthe probability that all three space girls, and also thenP(at least one young selected) = 1-P(all girl selected)to acquire the answer.The probability of picking a girl is P(girl selected first) = $\frac\textnumber of girl inclass\texttotal number in class= \frac1528$Then P(second selected additionally a girl) = $\frac1427$And P(third selected also a girl) = $\frac1326$So P(all girls selected) =$\frac1528\times\frac1427\times\frac1326 =\frac536$Then the answer isP(at least one young selected) = 1 - P(all girls selected) = 1 -$\frac536$ = $\frac3136$ Well done to everyone.
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