Probability Form 5

7/29/2019 Probability Form 5

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5 / 6 MARKS


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SAMPLE SPACE

An experiment is a process or an operationwith an outcomes.

Toss a balanced dice once and observe

its uppermost face.


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When toss the coin, we can get only2 results:

1. Head2. Tail


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The set of all possible outcomes of anexperiment is called the sample space.It usually denoted by S.

Example 1:

En. Adam has a fruit stall that sells bananas, apples,watermelons, papayas and durians. Students of class5M are asked to select their favorite fruit from thefruits at En. Adams stall.

Solution:S = { banana, apple, watermelon, papaya, durian}


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Example 2:

A month is randomly selected from ayear. Describe the sample space ofthis experiment by using set notation.

TOPIC MENU

Solution:

S= { January, February, March, April, May,

June, July, August, September, October,November, December}


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Sspace,samplein theoutcomespossibleofnumber

AeventforoutcomesofnumberP(A)

P(A) = 1 Event A is sure to happen

P(A) = 0 Event A will not to happen


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EXAMPLE 3 :

2 135 6 10 15

The above diagram shows six number cards in a box. Acard is picked at random from the box, find theprobability of obtaining

a) A prime number

b) A whole number

c) A multiple of 7

d) A number between 3 and 7


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Solution: Sample space , S = { 2, 5, 6, 10, 13, 15}

n(S) = 6

a) A prime number

(S)n(A)nP(A)

A = { 2, 5,13}

n(A) = 3

6

3P(A)

2

1

b) A whole number

(S)n

(A)nP(A)

A = { 2, 5, 6, 10, 13, 15}

n(A) = 6

6

6P(A)

1


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c) A multiple of 7 d) A number between 3and 7

(S)n

(A)nP(A)

A = { }

n(A) = 0

6

0

P(A)

0

Sample space ,

S = { 2, 5, 6, 10, 13, 15}n(S) = 6

A = { 5, 6 }n(A) = 2

(S)n

(A)nP(A)

6

2P(A)

3

1


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Calculating the probability of a combined event by listingthe outcomes.

The steps to calculate the probability of the combined events of

a) A or B b) A and B

Step 1 : List out all outcomes of the sample spaces.

Step 2 : List out all outcomes of the combined events.

a) A or B

Determine n (A B)

a) A and B

Determine n (A B)

Step 3 : Calculate.

)(

)(

Sn

BAn

)( BAP

)(

)(

Sn

BAn

)( BAP


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Calculating the probability of a combined event involvingthe sum of probability

A B

S

P (A B) = P(A) + P(B)

A B

S

P (A B) = P(A) + P(B) P(A B)


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Example 4 (SPM 2007)

The diagram shows ten labelled cards in two boxes.

A 2 B C D G3 E 4 F

A card is picked at random from each of the box.

By listing the outcomes, find the probability that

a) Both cards are labelled with a number. [ 2 marks ]

b) One card is labeled with a number and the other is labelled

with a letter. [ 2 marks ]


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Solution:

Step 1: List all the sample space

S = {

A 2 B C

D G3 E 4 F

(A,3), (A,E), (A,F), (A,G),(A,4),(A,D),

(2,3), (2,E), (2,F), (2,G),(2,4),(2,D),

(B,3), (B,E), (B,F), (B,G),(B,4),(B,D),

(C,3), (C,E), (C,F), (C,G),(C,4),(C,D),

1 mark

}


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}

a) Both cards are labelled with a number. [ 2 marks ]

S = { (A,3), (A,E), (A,F), (A,G),(A,4),(A,D),

(2,E), (2,F), (2,G),(2,4),(2,D),

(B,3), (B,E), (B,F), (B,G),(B,4),(B,D),

(C,3), (C,E), (C,F), (C,G),(C,4),(C,D),

(2,3),

(2,3)(2,4),={ }

24

2

12

1

A 2 B C

Method II

4

1

D G3 E 4 F 62

4

1

6

2

12

1


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S = { (A,3), (A,E), (A,F), (A,G),(A,4),(A,D),

(2,E), (2,F), (2,G),(2,4),(2,D),

(B,3), (B,E), (B,F), (B,G),(B,4),(B,D),

(C,3), (C,E), (C,F), (C,G),(C,4),(C,D),

(2,3),

b) One card is labeled with a number and the other is labelled

with a letter. [ 3 marks ]

(2,G),

(A,3),

(A,4), (2,E),

(2,F),(2,D),

(B,3), (B,4), (C,3),(C,4),

={

}

24

10

12

5 A 2 B C

Method II

D G3 E 4 F

6

4

4

1 +

6

2

4

3

12

5

}


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The diagram shows three numbered cards in box P andtwo cards in box Q.

2 3 6 Y R

A card is picked at random from box P and then a card ispicked at random from box Q

By listing the outcomes, find the probability that

a) A card with an even number and the card labelled Y are

picked. [ 2 marks ]

b) a card with a number which is multiple of 3 or the card

labelled R are picked. [ 3 marks ]


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Solution:

Step 1: List all the sample space

2 3 6 Y R

S = { (2, Y), (2, R), (3, Y), (3, R), (6, Y), (6, R)}

a) A card with an even number and the card labelled Y are picked.[ 2 marks ]

(2, Y), (6, Y),=

6

2

3

2

Method II

2 3 6 Y R

3

1

2

1

2

1

3

2

3

1


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S = { (2, Y), (2, R), (3, Y), (3, R), (6, Y), (6, R)}

b) a card with a number which is multiple of 3 or

the card labelled R are picked. [ 3 marks ]

)(

)(

Sn

BAn

)(or)( BPAP

Event A

Event B

)( BAP

6

5


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Method II

2 3 6 Y R3

2

2

1

)(or)( BPAP

)( BAP )()()( BAPBPAP

3

2 +

2

1

3

2x

2

1

3

1

6

34

6

5


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The diagram shows five cards labelled with letter.

S M I L E

All these cards are put into a box. A two-letter code is to beformed by using any of these cards. Two cards are picked atrandom, one after another, without replacement

a) List all the sample space [ 2 marks ]

b) List all the outcomes of the events and find the probability

that

i. The code begins with letter M,

ii. The code consists of two vowels or two consonents.

[ 4 marks ]


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Solution:

S = { ( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

M

S,

M,

L,

E,

I,

S,

M,

L,

E,

I,

S,

M,

L,

E,

I,

S,

M,

L,

E,

I,

I L E

S

S

S

S

I

M

M

M

M

I L

L E

L E

E

}

a) List all the sample space [ 2 marks ]


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b) List all the outcomes of the events and find the probability

that

i. The code begins with letter M [ 4 marks ]

S = { ( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

( ),

M

S,

M,

L,

E,

I,

S,

M,

L,

E,

I,

S,

M,

L,

E,

I,

S,

M,

L,

E,

I,

I L E

S

S

S

S

I

M

M

M

I

M M

L E

L E

E

}

20

4

5

1


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ii. The code consists of two vowels or two consonents.

S = { ( ),

( ),( ),

( ),

( ),

( ),

( ),( ),

( ),

( ),

( ),

( ),( ),

( ),

( ),

( ),

( ),( ),

( ),

( ),

M

S,

M,

L,

E,

I,

S,

M,

L,

E,

I,

S,

M,

L,

E,

I,

S,

M,

L,

E,

I,

I L E

SS

S

S

I

M

M

M

I

I M

L EL E

E

}

20

8

5

2

Method II

S M I L E

4

2

5

3+

4

1

5

2

5

2

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Probability Form 5