We got a big number of responses ofexcellent quality. Ben from St Peter"s adhered to the treediagram and also calculated out the answer:If you flip a coin three times the chance of acquiring at least onehead is 87.5%. To obtain this result I used the detailed treediagram to create how plenty of outcomes supplied one head.Llewellyn indigenous St Peter"s and also Diamor fromWillington ar Grammar college both it was observed an interestingpattern and expanded the answer to flipping ten coins:If you upper and lower reversal a coin 3 time the probability of gaining at the very least oneheads is 7 in 8 by analysis the table. This table additionally works theopposite way, the possibilities of Charlie gaining no top is 1 in 8because the end of every the outcomes only one of them has only tails. Inotice the if you add these probabilities with each other you get thetotal lot of outcomes (7+1=8). If you flip a coin 4 time theprobability the you obtaining at the very least one top is 15 in 16 becauseyou times the amount of outcomes friend can gain by flipping 3 coins by2, it outcomes in 16 and also then friend minus 1 indigenous it. V 5 coins toflip you just times 16 through 2 and also then minus 1, for this reason it would certainly resultwith a 31 in 32 chance of acquiring at least one heads. Through 6 coinsyou times by 2 and also minus by 1 again causing a 63 in 64 chance.To uncover the opportunity of getting at least one top if you flip tencoins you time 64 by 2 4 times or through 16 once and then minus 1,this results in a 1063 in 1064 chance of gaining at the very least oneheads.  Neeraj from Wilson School developed ageneralization for different numbers that possibleoutcomes: ns noticed that once you add the probabilities together they make awhole. A quick means of figuring out how countless times you acquire at leastone head is, that it is constantly the no. Of feasible outcomes minusone end the no. Of possible outcomes So: if No that possibleoutcomes = n the equation would be: P= (n- 1)/n  One student suggested how to calculation thenumber of desired outcomes:If the number 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 number of outcomes through atleast one head is $2^10-1 = 1024 - 1 = 1023$ Luke from Maidstone Grammar Schoolwent further to inspection the next component of thequestion:When there space 4 environment-friendly balls in the bag and also there space 6 red ballsthe probability of randomly picking a eco-friendly ball is 0.4($\frac25$) and the probability selecting a red sphere is 0.6($\frac35$).If a ball is selected and also thenreplaced the probability of picking a red sphere or a greenball is the same every time. As soon as 3 balls space picked withreplacement the probability of obtaining at the very least one environment-friendly is1-(the probability of getting 3 reds)Because the probability is the same every time the possibility ofgetting 3 reds is $0.6^3=0.216$ (or in fountain $(\frac35)^3 =\frac27125$). Therefore the probability of gaining at least one greenis $1-0.216=0.784$ (or in fractions $1 - \frac27125 =\frac98125$). When the balls space notreplaced the probability of acquiring at the very least one eco-friendly isstill 1-(the probability of gaining 3 reds). In each attract theprobability of drawing a red ball is $\frac\textthe variety of redballs\textthe total number of balls$On the first draw there space 6 red balls out of 10 so theprobability of choose a red is $\frac610$.On the 2nd draw there room 5 red balls the end of 9 therefore theprobability of choose a red is $\frac59$.On the final attract there space 4 red balls out of 8 therefore the probabilityof picking a red is $\frac48$.The probability the this succession of draws happening is theprobability that each attract multiplied together. I.e.:$\frac610\times\frac59\times\frac48=\frac16$The probability of drawing all reds is $\frac16$ and so theprobability of drawing at least one eco-friendly is $\frac56$. Helen indigenous Stroud finished up theproblem: When children are selected because that the institution council they room notreplaced.

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