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Conditional Probability: Dependent Events When events are not independent, the outcome of earlier events affects the outcome of later events. This happens in situations when the objects selected are not replaced.

A box of chocolates contains twelve chocolates of three different types. There are 3 strawberry, 4 caramel and 5 milk chocolates in the box. Sam chooses a chocolate at random and eats it. Jenny then does the same. Calculate the probability that they both choose a strawberry chocolate. Conditional Probability: Dependent Events P(strawberry and strawberry) =3/12 x

A box of chocolates contains twelve chocolates of three different types. There are 3 strawberry, 4 caramel and 5 milk chocolates in the box. Sam chooses a chocolate at random and eats it. Jenny then does the same. Calculate the probability that they both choose a strawberry chocolate. Conditional Probability: Dependent Events P(strawberry and strawberry) =3/12 x 2/11 = 6/132 (1/22)

Conditional Probability: Dependent Events P(milk and caramel) =5/12 x 1. Calculate the probability that Sam chooses a milk chocolate and Jenny chooses a caramel chocolate.

Conditional Probability: Dependent Events P(milk and caramel) = 1. Calculate the probability that Sam chooses a milk chocolate and Jenny chooses a caramel chocolate. 4/11 = 20/132 (5/33) 5/12 x

Conditional Probability: Dependent Events P(caramel and strawberry) =4/12 x 2. Calculate the probability that Sam chooses a caramel chocolate and Jenny chooses a strawberry chocolate.

Conditional Probability: Dependent Events P(caramel and strawberry) =4/12 x 2. Calculate the probability that Sam chooses a caramel chocolate and Jenny chooses a strawberry chocolate. 3/11 = 12/132 (1/11)

Conditional Probability: Dependent Events P(strawberry and not strawberry) =3/12 x 3. Calculate the probability that Sam chooses a strawberry chocolate and Jenny does not choose a strawberry chocolate.

Conditional Probability: Dependent Events P(strawberry and not strawberry) =3/12 x 3. Calculate the probability that Sam chooses a strawberry chocolate and Jenny does not choose a strawberry chocolate. 9/11 = 27/132 (9/44)

S C RABBL E E P T P M E E X A Jenny picks up two extra letters by mistake at the start of a game of scrabble. Sam removes two of the letters at random from Jenny without looking. Calculate the probability that he removes: (a) Two letter E’s P(E and E) = 3/9 x

S C RABBL E E P T P M E X A Jenny picks up two extra letters by mistake at the start of a game of scrabble. Sam removes two of the letters at random from Jenny without looking. Calculate the probability that he removes: (a) Two letter E’s P(E and E) = 3/9 x 2/8 = 6/72 (1/12)

S C RABBL E E P T P M E E X A Jenny picks up two extra letters by mistake at the start of a game of scrabble. Sam removes two of the letters at random from Jenny without looking. Calculate the probability that he removes: (b) A letter P followed by a letter E P(P and E) = 2/9 x

S C RABBL E E P T M E X A 3/8 = 6/72 (1/12) Jenny picks up two extra letters by mistake at the start of a game of scrabble. Sam removes two of the letters at random from Jenny without looking. Calculate the probability that he removes: (b) A letter P followed by a letter E P(P and E) = 2/9 x E

S C RABBL E E P T P M E E X A Jenny picks up two extra letters by mistake at the start of a game of scrabble. Sam removes two of the letters at random from Jenny without looking. Calculate the probability that he removes: (c) A letter X followed by a letter E P(X and E) = 1/9 x

S C RABBL E E P T M E A 3/8 = 3/72 (1/24) Jenny picks up two extra letters by mistake at the start of a game of scrabble. Sam removes two of the letters at random from Jenny without looking. Calculate the probability that he removes: (c) A letter X followed by a letter E P(X and E) = 1/9 x E P

S C RABBL E E P T P M E E X A Jenny picks up two extra letters by mistake at the start of a game of scrabble. Sam removes two of the letters at random from Jenny without looking. Calculate the probability that he removes: (d) A vowel followed by a letter P P(vowel and P) = 4/9 x2/8 = 8/72 = 1/9

S C RABBL E E P T P M E E X A Jenny picks up two extra letters by mistake at the start of a game of scrabble. Sam removes two of the letters at random from Jenny without looking. Calculate the probability that he removes: (e) A consonant followed by a letter E P(consonant and E) = 5/9 x3/8 = 15/72 (5/24)

Peter has ten coloured cubes in a bag. Three of the cubes are red and seven are blue. He removes a cube at random from the bag and notes the colour but does not replace it. He then removes a second cube at random. Calculate the probabilities for Peter selecting the following cubes: (a) A red followed by a blue: (b) Two reds: (c) Two blues: (d) A blue followed by a red: P(red and blue) = 3/10 x 7/9 = 21/90 P(red and red) = 3/10 x 2/9 = 6/90 P(blue and blue) = 7/10 x 6/9 = 42/90 P(blue and red) = 7/10 x 3/9 = 21/90

Rebecca has nine coloured beads in a bag as shown below. She removes a bead at random from the bag and does not replace it. She then chooses a second bead. Calculate the probabilities for Rebecca selecting the following beads: (a) A red followed by a blue: (b) Two reds: (c) Two blacks: (d) A blue followed by a red: P(red and blue) = 2/9 x 3/8 = 6/72 P(red and red) = 2/9 x 1/8 = 2/72 P(black and black) = 4/9 x 3/8 = 12/72 P(blue and red) = 3/9 x 2/8 = 6/72

Peter has ten coloured cubes in a bag. Three of the cubes are red and seven are blue. He removes a single cube at random from the bag and notes the colour but does not replace it. He repeats this for two more cubes. Calculate the probabilities for Peter selecting the following cubes in the order given: (a) Three reds: (b) Red, red, blue: (c) Blue, blue, red: (d) Three blues P(3 reds) = 3/10 x 2/9 x 1/8 = 6/720 P(red and red and blue) = 3/10 x 2/9 x 7/8 = 42/720 P(blue and blue and red) = 7/10 x 6/9 x 3/8 = 126/720 P(3 blues) = 7/10 x 6/9 x 5/8 = 210/720

Rebecca has nine coloured beads in a bag as shown below. She removes a bead at random from the bag and does not replace it. She then repeats this for two more beads. Calculate the probabilities for Rebecca selecting the following beads in the order given. (a) Three blacks: (b) Red, red, blue: (c) Blue, black, red: (d) Three blues P(3 blacks) = 4/9 x 3/8 x 2/7 = 24/504 P(red and red and blue) = 2/9 x 1/8 x 3/7= 6/504 P(blue and black and red) = 3/9 x 4/8 x 2/7 = 24/504 P(3 blues) = 3/9 x 2/8 x 1/7= 6/504

Peter has ten coloured cubes in a bag. Three of the cubes are red and seven are blue. He removes a cube at random from the bag and notes the colour but does not replace it. He then removes a second cube at random. Calculate the probability that Peter selects a red and a blue cube in any order. P(red and blue or blue and red) = 3/10 x 7/9 + 7/10 x 3/9 = 21/ /90 = 42/90

Rebecca has nine coloured beads in a bag as shown below. She removes a bead at random from the bag and does not replace it. She then chooses a second bead. Calculate the probabilities for Rebecca selecting the following beads in any order. (a) A red and a blue: (b) A blue and a black P(red and blue or blue and red) = 2/9 x 3/8 + 3/9 x 2/8 = 6/72 + 6/72 = 12/72 P(blue and black or black and blue) = 3/9 x 4/8 + 4/9 x 3/8 = 12/ /72 = 24/72 (c) A black and a red P(black and red or red and black) = 4/9 x 2/8 + 2/9 x 4/8 = 8/72 + 8/72 = 16/72