We obtained a large number of responses ofwonderful high quality. Ben from St Peter"s followed the treediagram and also calculated out the answer:If you flip a coin three times the chance of getting at least onehead is 87.5%. To get this outcome I provided the gave treediagram to create how many kind of outcomes provided one head.Llewellyn from St Peter"s and also Diamor fromWillington County Grammar School both oboffered an interestingpattern and expanded the answer to flipping ten coins:If you flip a coin 3 times the probcapacity of getting at least oneheads is 7 in 8 by analysis the table. This table additionally functions theoppowebsite means, the chances of Charlie acquiring no heads is 1 in 8bereason out of all the outcomes just one of them has actually only tails. Inotification that if you add these probabilities together you gain thecomplete amount of outcomes (7+1=8). If you flip a coin 4 times theprobcapability of you gaining at least one heads is 15 in 16 becauseyou times the amount of outcomes you can obtain by flipping 3 coins by2, it results in 16 and then you minus 1 from it. With 5 coins toflip you simply times 16 by 2 and then minus 1, so it would certainly resultvia a 31 in 32 chance of obtaining at leastern one heads. With 6 coinsyou times by 2 and also minus by 1 aget leading to a 63 in 64 opportunity.To discover the opportunity of gaining at least one heads if you flip tencoins you times 64 by 2 4 times or by 16 when and also then minus 1,this outcomes in a 1063 in 1064 possibility of acquiring at leastern oneheads. Neeraj from Wilson School emerged ageneralization for various numbers of possibleoutcomes: I noticed that once you add the probabilities together they make aentirety. A quick means of figuring out just how many type of times you get at leastone head is, that it is always the no. Of feasible outcomes minusone over the no. of possible outcomes So: if No of possibleoutcomes = n the equation would certainly be: P= (n- 1)/n One student suggested exactly how to calculate thenumber of preferred outcomes:If the number of times flipped =p Then the variety of outcomes thatcontain a head is$2^p-1$So for flipping a coin $10$ times, the number of outcomes via atleastern one head is $2^10-1 = 1024 - 1 = 1023$ Luke from Maidstone Grammar Schoolwent better to investigate the next component of thequestion:When tbelow are 4 green balls in the bag and tbelow are 6 red ballsthe probcapacity of randomly selecting a green ball is 0.4($frac25$) and also the probcapacity selecting a red ball is 0.6($frac35$).If a round is schosen and thenreplaced the probcapability of picking a red sphere or a greensphere is the very same eextremely time. When 3 balls are picked withreplacement the probcapacity of gaining at leastern one green is1-(the probability of getting 3 reds)Because the probability is the same eincredibly time the opportunity ofacquiring 3 reds is $0.6^3=0.216$ (or in fractions $(frac35)^3 =frac27125$). So the probcapacity of getting at least one greenis $1-0.216=0.784$ (or in fractions $1 - frac27125 =frac98125$). When the balls are notreplaced the probcapability of obtaining at leastern one green isstill 1-(the probcapability of getting 3 reds). In each draw theprobability of illustration a red round is $frac extthe number of redballs extthe total variety of balls$On the first attract tright here are 6 red balls out of 10 so theprobcapacity of picking a red is $frac610$.On the second attract there are 5 red balls out of 9 so theprobcapacity of picking a red is $frac59$.On the last draw tbelow are 4 red balls out of 8 so the probabilityof picking a red is $frac48$.The probcapability of this sequence of draws happening is theprobcapacity of each attract multiplied together. i.e.:$frac610 imesfrac59 imesfrac48=frac16$The probcapability of illustration all reds is $frac16$ and so theprobability of illustration at least one green is $frac56$. Helen from Stroud finished up theproblem: When children are selected for the institution council they are notreinserted.

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The youngsters are schosen one after an additional and each timethe probcapability of a boy being selected isP(boy schosen first) = $frac extthe variety of boys in theclass extthe full variety of kids in the class$Note: the class describes students that have actually not currently been madepart of the council.To find the probcapacity that there will certainly be at least one boy, findthe probability that all 3 are girls, and also thenP(at leastern one boy selected) = 1-P(all girls selected)to gain the answer.The probcapacity of picking a girl is P(girl schosen first) = $frac extnumber of girls inclass exttotal number in class= frac1528$Then P(second schosen also a girl) = $frac1427$And P(3rd selected likewise a girl) = $frac1326$So P(all girls selected) =$frac1528 imesfrac1427 imesfrac1326 =frac536$Then the answer isP(at leastern one boy selected) = 1 - P(all girls selected) = 1 -$frac536$ = $frac3136$ Well done to everyone.

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