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+ Probability: Independent Events 1 2 3 4 5 6 2 3 4 5 6 7 8 9 10 11 12 Consider the probability of throwing a double six with two dice. P(6 and 6) We could make a table of all possible outcomes and count the ones that we want. 1 2 3 4 5 6 + Complete the remainder of the table of outcomes. 2 3 4 5 6 7 8 9 10 11 12 3 4 4 5 5 6 6 7 P( 6 and 6) = 1/36 7 8

+ Probability: Independent Events 1 2 3 4 5 6 7 8 9 10 11 12 The throwing of 2 dice are independent events. This means that the outcomes on one die are not affected in any way by the outcomes on the other. 1 2 3 4 5 6 + 7 8 9 10 11 12 On a single throw of a die P(6) = 1/6 For 2 dice the probability of 2 sixes is the same as 1/6 x 1/6 = 1/36 Can you see why multiplying the individual probabilities together for one event, (rolling a die) gives us the correct results for 2 events, (rolling 2 dice)? This is because for each outcome on one die there are 6 outcomes on the other and so there are 6 x 6 = 36 outcomes in total.

1 1 4 2 3 4 5 4 2 4 Spinners Probability: Independent Events Consider the probability of getting a 2 on the blue spinner and a 4 on the red spinner. Would multiplying the probabilities 2/5 x 3/6 = 6/30 give the correct answer? 1 1 2 3 4 4 2 Spinners 5 4 4

1 2 2 + 1 4 5 3 4 Probability: Independent Events 1 2 3 4 5 3 2 5 6 4 Consider the probability of getting a 3 on the blue spinner and a 4 on the red spinner. Probability: Independent Events Would multiplying the probabilities 2/5 x 3/6 = 6/30 gives the correct answer? 2 3 1 4 Construct a table of outcomes to see whether or not this is correct. 4 2 5 1 1 2 3 4 5 + 3 2 5 6 4 7 8 9 A probability of 6/30 is correct. Notice again that the events are independent of each other.

For independent events A and B, P(A and B) = P(A) x P(B) Probability: Independent Events AND LAW The AND LAW For independent events A and B, P(A and B) = P(A) x P(B) P(6 AND 6) P(3 BLUE AND 4 RED) = 1/6 X 1/6 = 1/36 = 2/5 X 3/6 = 6/30 So rather than constructing tables of outcomes, or making lists of outcomes, we can simply multiply together the probabilities for the individual events.

For mutually exclusive events: P(A and B) = P(A) x P(B) The AND LAW When using the And Law and multiplying the individual probabilities, the cumulative effect decreases the likelihood of the combined events happening. 3 2 1 ½ 1 Impossible Certain P(6) = 1/6 P(double 6) = 1/36 Common sense should tell you that it is much harder (less likely) to throw a double six with two dice, then it is to throw a single 6 with one die. What do you think the probability of getting a triple 6 with a throw of 3 dice might be? P(triple 6) 1/6 x 1/6 x 1/6 = 1/216

For mutually exclusive events: The AND LAW For mutually exclusive events: P(A and B) = P(A) x P(B) Impossible 1 Certain ½ When using the And Law and multiplying the individual probabilities, the cumulative effect decreases the likelihood of the combined events happening. P(6) = 1/6 P(double 6) = 1/36 2 3 Common sense should tell you that it is much harder (less likely) to throw a double six with two dice, then it is to throw a single 6 with one die. This reduction may seem obvious but it is worth stating explicitly since there is often confusion with the OR LAW (done earlier) where the probabilities increase. Impossible 1 Certain ½ P(red) = 3/12 P(red or blue ) = 3/12 + 5/12 = 8/12 P(red or blue or black ) = 3/12 + 5/12 + 4/12 = 1 2 3 OR LAW

For independent events A and B, P(A and B) = P(A) x P(B) The AND LAW Probability: Independent Events Both players lay a card at random from their hands as shown. Question 1. What is the probability that two kings are laid? Player 2 Player 1 P(king and king) = 2/7 x 1/7 = 2/49 Cards

For independent events A and B, P(A and B) = P(A) x P(B) The AND LAW Probability: Independent Events Both players lay a card at random from their hands as shown. Question 2. What is the probability that two hearts are laid? Player 1 Player 2 P(heart and heart) = 4/6 x 2/8 = 8/48 (1/6)

The AND LAW Probability: Independent Events P(picture and picture) = For independent events A and B, P(A and B) = P(A) x P(B) The AND LAW Probability: Independent Events Both players lay a card at random from their hands as shown. Q 3. What is the probability that two picture cards are laid? (Ace is not a picture card) Player 1 Player 2 P(picture and picture) = 2/8 x 5/7 = 10/56 (5/28)

Beads (a) P(red and red) = 4/9 x 4/9 = 16/81 (b) P(blue and red) = Question 4. Rebecca has nine coloured beads in a bag. Four of the beads are red and the rest are blue. She removes a bead at random from the bag and notes the colour before replacing it. She then chooses a second bead. Calculate the probability that Rebecca chooses: (a) 2 red beads (b) A blue followed by a red bead. Beads (a) P(red and red) = 4/9 x 4/9 = 16/81 (b) P(blue and red) = 5/9 x 4/9 = 20/81

(a) P(red and red) = 3/8 x 6/9 = 18/72 = 1/4 (b) P(blue and blue) = Question 5. Peter and Rebecca each have a bag of red and blue beads as shown below. They each remove a bead at random from their bags. Peter selects his bead first. Calculate the probability that: (a) Both beads will be red (b) Both beads will be blue (c) Peter’s bead is red and Rebecca’s is blue. (answers in simplest form) (a) P(red and red) = 3/8 x 6/9 = 18/72 = 1/4 (b) P(blue and blue) = 5/8 x 3/9 = 15/72 = 5/24 (c) P(red and blue) = 3/8 x 3/9 = 9/72 = 1/8

5 1 2 2 3 4 5 2 2 4 Spinners Probability: Independent Events The pointers on both spinners shown below are spun. Calculate the following probabilities: (a) A 2 on both spinners. (b) A 2 on the blue spinner and a 5 on the red spinner. 5 1 2 3 4 2 2 5 2 4 (a) P(2 and 2) = 2/5 x 3/6 = 6/30 (1/5) (b) P(2 and 5) = 2/5 x 2/6 = 4/30 (2/15)

6 5 4 8 4 3 Probability: Independent Events (a) P(4 and 4) = The pointers on both spinners shown below are spun. Calculate the following probabilities: (a) A 4 on both spinners. (b) A 5 on the red spinner and an 8 on the blue spinner. 3 4 8 4 6 5 (a) P(4 and 4) = 2/6 x 3/8 = 6/48 (1/8) (b) P(5 and 8) = 3/6 x 3/8 = 9/48 (3/16)

2 4 6 9 3 8 1 5 Probability: Independent Events (a) P(4 and 5) = The pointers on the spinners are spun in the order shown. Calculate the probabilities for the outcomes shown on the following pairs of spinners : (a) Pink and red (b) Pink and blue (c) Red and blue 4 1 2 3 5 6 8 9 (a) P(4 and 5) = 2/5 x 3/6 = 6/30 (1/5) (b) P(4 and 6) = 2/5 x 3/8 = 6/40 (3/20) (c) P(5 and 6) = 3/6 x 3/8 = 9/48 (3/16)

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