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Probability: Independent Events Consider the probability of throwing a double six with two dice. P(6 and 6) Complete the remainder of the table of outcomes. P( 6 and 6) = 1/36 We could make a table of all possible outcomes and count the ones that we want.

Probability: Independent Events 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. 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.

Spinners Consider the probability of getting a 2 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 give the correct answer?

Construct a table of outcomes to see whether or not this is correct 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? Notice again that the events are independent of each other A probability of 6/30 is correct.

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

The AND LAW For mutually exclusive events: P(A and B) = P(A) x P(B) 0 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. 1 P(6) = 1/6 P(double 6) = 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. 2 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/ /36

The AND LAW For mutually exclusive events: P(A and B) = P(A) x P(B) 0 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. 1 P(6) = 1/6 P(double 6) = 1/36 2 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. 0 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 = OR LAW 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.

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

Player 1 Player 2 P(heart and heart) = 4/6 x 2/8 = 8/48 (1/6) Probability: Independent Events For independent events A and B, P(A and B) = P(A) x P(B) The AND LAW 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 P(picture and picture) = 2/8 x 5/7 = 10/56 (5/28) Probability: Independent Events For independent events A and B, P(A and B) = P(A) x P(B) The AND LAW 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 2

Beads 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. (a) P(red and red) = 4/9 x 4/9 = 16/81 (b) P(blue and red) = 5/9 x 4/9 = 20/81

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

The pointers on both spinners shown below are spun. Probability: Independent Events Calculate the following probabilities: (a) A 2 on both spinners. (b) A 2 on the blue spinner and a 5 on the red spinner. (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) Spinners

The pointers on both spinners shown below are spun. Probability: Independent Events Calculate the following probabilities: (a) A 4 on both spinners. (b) A 5 on the red spinner and an 8 on the blue spinner. (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)

The pointers on the spinners are spun in the order shown. Probability: Independent Events 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 (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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