YEAR 11 MATHS REVISION Probability

Probability from a table So, all we have to do is multiply the probability by the total……. 0.25 x 300 = Probabilities must always add up to 1, so…….

Probability from a table So, all we have to do is multiply the probability by the total……. 0.2 x 240 Probabilities must always add up to 1, so…….

Probability from a table

Probability from a table

Probability tree diagrams 100 9 100 49 100 58 P(The same colour) = 50 29 Hubert has 10 coloured counters in a bag. Three of the counters are blue and seven of the counters are green. He removes a counter at random from the bag and notes the colour before replacing it. He then chooses a second counter. Record the information on a tree diagram and work out the probabilities of the different outcomes. P(B,B) + P(G,G) = + = = 100 21 100 21 100 42 50 21 P(BG in any order) = P(B,G) + P(G,B) = + = = FIRST PICK SECOND PICK ALWAYS MULTIPLY THE BRANCHES 10 3 10 3 100 9 3 10 Blue P(B,B) = X = 3 10 Blue COUNTERS HAVE BEEN REPLACED SO THE PROBABILITY DOES NOT CHANGE ON THE SECOND PICK 7 10 10 3 10 7 100 21 Green P(B,G) = X = 10 7 10 3 100 21 3 10 Blue P(G,B) = X = 7 10 Green 7 10 10 7 10 7 100 49 Green P(G,G) = X =

Probability tree diagrams Peter and Becky run a race and play a tennis match. The probability that Peter wins the race is 0.4. The probability that Becky wins the tennis is 0.7. (a) Complete the tree diagram below. (b) Use your tree diagram to calculate (i) the probability that Peter wins both events. (ii) The probability that Becky loses the race but wins at tennis. Race Tennis 0.6 0.3 0.7 Peter Win P(Win and Win) for Peter = 0.12 0.4 x 0.3 = 0.12 0.4 x 0.7 = 0.28 0.6 x 0.3 = 0.18 0.6 x 0.7 = 0.42 Peter Win 0.4 Becky Win P(Lose and Win) for Becky = 0.28 0.7 Peter Win Becky Win Becky Win

Probability tree diagrams 90 6 90 42 90 48 P(The same colour) = 15 8 Hubert has 10 coloured counters in a bag. Three of the counters are blue and seven of the counters are green. He removes a counter at random from the bag and notes the colour and does not replace it. He then chooses a second counter. Record the information on a tree diagram and work out the probabilities of the different outcomes. P(B,B) + P(G,G) = + = = 90 21 90 21 90 42 15 7 P(BG in any order) = P(B,G) + P(G,B) = + = = FIRST PICK SECOND PICK ALWAYS MULTIPLY THE BRANCHES 10 3 9 2 90 6 2 9 Blue P(B,B) = X = 3 10 Blue COUNTERS HAVE NOT BEEN REPLACED SO THE PROBABILITY DOES CHANGE ON THE SECOND PICK 7 9 10 3 9 7 90 21 Green P(B,G) = X = 10 7 9 3 90 21 3 9 Blue P(G,B) = X = 7 10 Green 6 9 10 7 9 6 90 42 Green P(G,G) = X =