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SLIDE 1
Revision
? There are 80 children in a class of which 43 are girls. If one student is chosen at random, what is the probability that the student will be a boy?
? There are 52 cards in a pack of playing cards. If one card is selected at random, what is the probability of that card being either one of the Kings or the Queen of Hearts?
SLIDE 2
? There are 80 children in a class of which 43 are girls. If one student is chosen at random, what is the probability that the student will be a boy?
? There are 52 cards in a pack of playing cards in four suits. If one card is selected at random, what is the probability of that card being either one of the Kings or the Queen of Hearts?
SLIDE 3
Revision
Law of total probability
If all the events are given, then the sum of the probability of all the events is 1.
Question
There are only blue counters, green counters, red counters and black counters in a bag. The table shows the probability of getting a blue counter or a green counter or a red counter. (a) What is the probability of getting a black counter? (b) What is the least possible number of counters in the bag?
Colour blue green red black
Probability 0.3 0.35 0.25
SLIDE 4
The probability of a black counter is
Colour blue green red black
Probability 0.3 0.35 0.25 0.1
Least counters 6 7 5 2
There are fractions of 0.05 in the table. As the probability is there must be at least 20 counters in the bag.
SLIDE 5
A game uses tiles which are coloured blue, yellow or green only. There are three times as many blue tiles as yellow tiles and five times as many green tiles as blue tiles. If a single tile is taken at random what is the probability that the tile is green?
SLIDE 6
A game uses tiles which are coloured blue, yellow or green only. There are three times as many blue tiles as yellow tiles and five times as many green tiles as blue tiles. If a single tile is taken at random what is the probability that the tile is green?
Solution
The least numbers of tiles are
blue : yellow : green
Total number of tiles
Then the probability of a green counter is
SLIDE 7
Two dice are rolled, and the scores are multiplied together.
Complete the table
Number of first die
Number of second die 1 2 3 4 5 6
1
2
3
4
5
6
Use the table to find the probability when rolling two dice of obtaining a score of (a) 4, (b) 25, (c) an odd number.
SLIDE 8
Number of first die
Number of second die 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 8 10 12
3 3 6 9 12 15 18
4 4 8 12 16 20 24
5 5 10 15 20 25 30
6 6 12 18 24 30 36
There are 36 possible outcomes.
(a) Three outcomes have a score of 4.
(b) Only one outcome is 25
(b) Nine outcomes are odd
SLIDE 9
Example
A fair coin is spun twice. Draw a probability tree and find the probability of obtaining one head.
Solution
There are two ways of obtaining one head: either head first, then tail, or tail first, then head.
SLIDE 10
The diagram shows how to calculate probabilities in a probability tree. As we proceed along a branch we multiply the probabilities. If we have to combine outcomes of branches, as in this example, we add the probabilities of each branch.
Here the value of each branch is , and as there are two branches giving an outcome of one head, we add the probability of each: .
SLIDE 11
A fair dice is thrown twice.
Complete the above probability tree and find the probability of obtaining an odd and an even number in any order.
SLIDE 12
SLIDE 13
A fair coin is spun three times. Draw a probability tree and find the probability of obtaining two heads.
SLIDE 14
A fair coin is spun three times. Draw a probability tree and find the probability of obtaining two heads.
The probability of H T H in that order is found by multiplying the probabilities along the branches: . Here every branch has the same probability. There are three ways of obtaining just two heads:
H H T H T H T H H
The probability of two heads is .
SLIDE 15
Use the probability tree to complete the following probability table
number of heads 0 1 2 3
number of possibilities
probability
SLIDE 16
number of heads 0 1 2 3
number of possibilities 1 3 3 1
probability
SLIDE 17
With and without replacement
Suppose a bag contains 12 counters and 5 are black and 7 are white. When you take one counter out of the bag there are two possible situations: either you put the counter back into the bag, or you leave the counter out of the bag. The first case is called with replacement and the second is called without replacement.
? With replacement
Suppose you take a black counter out of the bag, and put it back in. What is the probability that the second counter will be black?
? Without replacement
Suppose you take a black counter out of the bag, and do not put it back in. What is the probability that the second counter will be black?
SLIDE 18
Suppose a bag contains 12 counters and 5 are black and 7 are white.
? With replacement
Suppose you take a black counter out of the bag, and put it back in. What is the probability that the second counter will be black?
The probability of the first counter being black is . If you put the counter back, then the probability that the second counter is black is exactly the same – in this case, .
? Without replacement
Suppose you take a black counter out of the bag, and do not put it back in. What is the probability that the second counter will be black?
The probability of the first counter being black is . If you do not put the counter back, then the probability that the second counter is black is altered. There are now only 11 counters in the bag in total, and of these 4 are black. So, the probability that the second counter is black is .
SLIDE 19
A bag contains 12 counters: 5 of these are black and 7 are white. Two counters are removed from the bag. By drawing probability trees (a) with replacement, (b) without replacement, find the probability of obtaining a black and a white counter in any order.
SLIDE 20
A bag contains 12 counters: 5 of these are black and 7 are white. Two counters are removed from the bag. By drawing probability trees (a) with replacement, (b) without replacement, find the probability of obtaining a black and a white counter in any order.
Solution
With replacement
Without replacement
SLIDE 21
Spinners A and B are spun. Each spinner can land only on blue or green. The probability that spinner A will land on blue is 0.8. The probability that spinner B will land on green is 0.25. Each spinner is spun once. The number of times both spinners land on blue is 36. Complete the probability tree below and estimate how many times both spinners landed on green.
In the diagram, the events for the different spinners indicated by the headings Spinner A and Spinner B.
SLIDE 22
The total number of trials in the experiment was .
SLIDE 23
A bag has 8 red, 5 yellow and 3 green apples. Two apples were taken at random from the bag and eaten. What is the probability that they were different colours?
Hint. Since the apples were eaten, this must be without replacement.
SLIDE 24
A bag has 8 red, 5 yellow and 3 green apples. Two apples were taken at random from the bag and eaten. What is the probability that they were different colours?
Solution
To get different colours we are require red followed by not red, yellow followed by not-yellow and green followed by not-green.
SLIDE 25
A team has two successive matches. The probability that it will win the first is 0.55. If the team wins the first, the probability it will win the second is 0.6. If it does not win the first, the probability it will win the second is 0.3. What is the probability that the team will win exactly one of its two matches?
SLIDE 26
A team has two successive matches. The probability that it will win the first is 0.55. If the team wins the first, the probability it will win the second is 0.6. If it does not win the first, the probability it will win the second is 0.3. What is the probability that the team will win exactly one of its two matches?
Solution
SLIDE 27
The probability that Bill will travel to work by car, motorcycle or train on a randomly chosen day is , and respectively. The probability of being late in each case is , and respectively. Find the probability that on a randomly chosen day Bill is not late.
SLIDE 28
The probability that Bill will travel to work by car, motorcycle or train on a randomly chosen day is , and respectively. The probability of being late in each case is , and respectively. Find the probability that on a randomly chosen day Bill is not late.
Solution
SLIDE 29
There are three machines in a factory A, B and C. They produce metal rods of the same length. Machine A produces 20% of the rods, machine B, 35% and machine C produces the rest. Machines A, B and C produce 2%, 5% and 10% defective rods respectively. Draw a tree diagram to represent this information and find the probability that a randomly selected rod is defective.
SLIDE 30
There are three machines in a factory A, B and C. They produce metal rods of the same length. Machine A produces 20% of the rods, machine B, 35% and machine C produces the rest. Machines A, B and C produce 2%, 5% and 10% defective rods respectively. Draw a tree diagram to represent this information and find the probability that a randomly selected rod is defective.
Solution
SLIDE 31
At a factory that produces toy teddy bears, the bears may be defective in three ways: they can have defective stitching, they can split open, and they can fade. The probability of a toy bear having defective stitching is 0.02. If the bear has defective stitching, the probability of it splitting open is 0.5, but a bear without defective stitching has a probability of just 0.05 of splitting open. The probability of a toy teddy bear fading is independent of the other defects and is 0.05. Find the probability that (a) a randomly chosen bear has none of the three defects, (b) that a randomly chosen bear has exactly one of the three defects.
Independent events
In this question, saying that the probability of the bear fading is independent of the other defects means that its probability is not altered by whether the bear does or does not have the other two defects.
SLIDE 32
At a factory that produces toy teddy bears, the bears may be defective in three ways: they can have defective stitching, they can split open, and they can fade. The probability of a toy bear having defective stitching is 0.02. If the bear has defective stitching, the probability of it splitting open is 0.5, but a bear without defective stitching has a probability of just 0.05 of splitting open. The probability of a toy teddy bear fading is independent of the other defects and is 0.05. Find the probability that (a) a randomly chosen bear has none of the three defects, (b) that a randomly chosen bear has exactly one of the three defects.
Independent events
In this question, saying that the probability of the bear fading is independent of the other defects means that its probability is not altered by whether the bear does or does not have the other two defects.
SLIDE 32
At a factory that produces toy teddy bears, the bears may be defective in three ways: they can have defective stitching, they can split open, and they can fade. The probability of a toy bear having defective stitching is 0.02. If the bear has defective stitching, the probability of it splitting open is 0.5, but a bear without defective stitching has a probability of just 0.05 of splitting open. The probability of a toy teddy bear fading is independent of the other defects and is 0.05. Find the probability that (a) a randomly chosen bear has none of the three defects, (b) that a randomly chosen bear has exactly one of the three defects.
Solution
SLIDE 33
The bag X contains six counters: 3 are white and 3 are black. The bag Y contains 5 counters, 2 are white and 3 are black. A counter is drawn at random from bag X and placed in bag Y. Another counter is then drawn from bag X and placed in bag Y. Bag Y now has seven counters in it. Then a counter is drawn from bag Y. Find the probability that the counter taken from bag Y is white.
SLIDE 34
The bag X contains six counters: 3 are white and 3 are black. The bag Y contains 5 counters, 2 are white and 3 are black. A counter is drawn at random from bag X and placed in bag Y. Another counter is then drawn from bag X and placed in bag Y. Bag Y now has seven counters in it. Then a counter is drawn from bag Y. Find the probability that the counter taken from bag Y is white.
Solution
The first two columns of the diagram show the result of picking the two counters from the first bag X. The last column is the counter drawn from bag Y. The number of white and black counters in bag Y depends on what counters were drawn from bag X.
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