PROBABILITY TREES
Probability trees are a method of visually calculating probabilities by constructing a sequence of events. They show all the possible outcomes of an experiment like the results of tossing a coin twice (shown below). The red line separates the first flip from the second flip.
Each branch of the tree represents a possible outcome. The fractions along the branches represent the probabilities of that branch. H = heads and T =tails. It works as follows:
In the first flip there are two outcomes, either H or T. If the result is H then follow the top branch. If not follow the bottom branch. The second flip results in either H or T, if H then follow that branch or if T, follow that branch. You multiply the fractions along the path or branch you follow. So HH = 1/2 x 1/2 or 1/4. This means the chance of getting HH is 0.25 or 25% or 1/4.
The probability of TT is similar and by following the lower branches you get TT also having probabilities of 0.25 or 25% OR 1/4.
The probability of getting a H and a T can be found looking at the outcome for TH and HT. Since this occurs on two occasions the probabilities are added together. So the probability of getting an H and a T is 1/4 + 1/4 which equals 0.5 or 50% or 1/2. From this we can say the probability of getting a T and a H is greater than that of getting HH.
In the first flip there are two outcomes, either H or T. If the result is H then follow the top branch. If not follow the bottom branch. The second flip results in either H or T, if H then follow that branch or if T, follow that branch. You multiply the fractions along the path or branch you follow. So HH = 1/2 x 1/2 or 1/4. This means the chance of getting HH is 0.25 or 25% or 1/4.
The probability of TT is similar and by following the lower branches you get TT also having probabilities of 0.25 or 25% OR 1/4.
The probability of getting a H and a T can be found looking at the outcome for TH and HT. Since this occurs on two occasions the probabilities are added together. So the probability of getting an H and a T is 1/4 + 1/4 which equals 0.5 or 50% or 1/2. From this we can say the probability of getting a T and a H is greater than that of getting HH.
probability with replacement
This means that the probabilities will always be out of the total number of objects in question because they are put back before the next draw. The following example uses shuffled cards but any situation like coloured jelly beans from a jar would be just the same.
A small pack of cards, 2 red and 3 black are shuffled and two cards are drawn at random. The results are recorded and the cards returned to the pack (replacement). What is the probability the two cards are red? Firstly a probability tree can help list the possible outcomes.
The possible outcomes are:
0 red cards (BB) which has a probability of 9/25 (the bottom tree branches)
1 red card (RB and BR) which has a probability of 6/25 + 6/25 = 12/25 (the two middle branches)
2 red cards (RR) which has a probability of 4/25 (the top tree branches)
The fractions at the end of the branches add to give 1 that is: 4/25 + 6/25 + 6/25 + 9/25 = 25/25 = 1
Some texts might describe the outcomes of this activity as:
0 red cards (NN - this means the two red cards are not red)
1 red card (RN and NR - this means 1 red card and 1 card that's not red)
2 red cards (RR - two red cards as before).
0 red cards (BB) which has a probability of 9/25 (the bottom tree branches)
1 red card (RB and BR) which has a probability of 6/25 + 6/25 = 12/25 (the two middle branches)
2 red cards (RR) which has a probability of 4/25 (the top tree branches)
The fractions at the end of the branches add to give 1 that is: 4/25 + 6/25 + 6/25 + 9/25 = 25/25 = 1
Some texts might describe the outcomes of this activity as:
0 red cards (NN - this means the two red cards are not red)
1 red card (RN and NR - this means 1 red card and 1 card that's not red)
2 red cards (RR - two red cards as before).
This terminology is useful when more complicated outcomes are encountered for example, you have 10 coloured cards in a pack, 6 red, 3 blue and 1 green, (see tree diagram above). What is the probability of taking out two red cards at random? (The cards are returned to the pack before the next selection 'replacement'). This gives 6/10 for red, and 3/10 for blue followed by 1/10 for green. However, it is easier to work with two variables rather than three. So blue and green can be combined to give 4/10 that are not red or 'N')
The possible outcomes are: (two card draw)
0 red cards, represented by NN
1 red card, represented by YN and NY
2 red cards, represented by YY.
0 red cards, represented by NN
1 red card, represented by YN and NY
2 red cards, represented by YY.
Once again:
0 red cards (NN) which has a probability of 16/100 (the bottom tree branches)
1 red card (RN and NR) which has a probability of 24/100 + 24/100 = 48/100 (the two middle branches)
2 red cards (RR) which has a probability of 36/100 (the top tree branches)
The total fractions add to give 16/100 + 24/100 + 24/100 + 36/100 = 100/100 = 1
0 red cards (NN) which has a probability of 16/100 (the bottom tree branches)
1 red card (RN and NR) which has a probability of 24/100 + 24/100 = 48/100 (the two middle branches)
2 red cards (RR) which has a probability of 36/100 (the top tree branches)
The total fractions add to give 16/100 + 24/100 + 24/100 + 36/100 = 100/100 = 1
probability without replacement
Probability without replacement means that the objects are not returned to the 'box, jar or bag'. So the total number of cards decreases by one after each draw. That is why in the second branch, the fraction denominators are now 9 and not 10. This will affect the probabilities compared to probability with replacement.
Assume a red card is drawn first. There are now 5 red cards and 4 'other cards' (not red cards) giving a total of 9 cards in the pack. So the probability of drawing another red card is now 5/9 and another card now 4/9 (still 4 'other cards' but 9 in total).
Assume a red card was not chosen first (follow the lower branches). There is a 4/10 chance of that happening. For the second draw there are now 3 'other cards' and 9 in total. The probabilities are as follows:
0 red cards (NN 12/90 from the bottom branches)
1 red card (RN and NR 24/90 + 24/90 from the two middle branches)
2 red cards (RR 30/90 from the top branches)
This adds to 12/90 + 24/90 + 24/90 + 30/90 = 90/90 = 1
Assume a red card was not chosen first (follow the lower branches). There is a 4/10 chance of that happening. For the second draw there are now 3 'other cards' and 9 in total. The probabilities are as follows:
0 red cards (NN 12/90 from the bottom branches)
1 red card (RN and NR 24/90 + 24/90 from the two middle branches)
2 red cards (RR 30/90 from the top branches)
This adds to 12/90 + 24/90 + 24/90 + 30/90 = 90/90 = 1
three draws probability without replacement
When three draws are carried out there are three branches to the tree diagram. We will not look at any branches greater than this. Using the same example as above, calculate the probability of drawing three red cards from a pack of 10.
Firstly, notice after each draw how the total reduces by 1. We started with 10 cards. so the third card draw is out of 8. There is also an easier way to determine the probabilities using patterns shown in the tree. But it's nice to have tree drawn because all the calculations are visible. There are four major patterns to identify and the first two are easy.
The first two patterns are three red cards and no red cards because you just multiply along the top branches and bottom branches, all the numbers reduce by one as you go along. The tricky ones are the inside branches.
The first two patterns are three red cards and no red cards because you just multiply along the top branches and bottom branches, all the numbers reduce by one as you go along. The tricky ones are the inside branches.
RRR = 120/720 from the upper branch
NNN = 24/720 from the lower branch
now the inner branches where there are two reds
RRN (each of these are 120/720 so that's 3 x 120/720)
RNR
NRR
Now the inner branches where there is one red
RNN (each of these are 72/720 so that's 3 x 72/720)
NRN
NNR
The total fractions add to (120/720 + 24/720 + 3(120/720) + 3(72/720)= 720/720 = 1
NNN = 24/720 from the lower branch
now the inner branches where there are two reds
RRN (each of these are 120/720 so that's 3 x 120/720)
RNR
NRR
Now the inner branches where there is one red
RNN (each of these are 72/720 so that's 3 x 72/720)
NRN
NNR
The total fractions add to (120/720 + 24/720 + 3(120/720) + 3(72/720)= 720/720 = 1