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Apply Advanced Conditional Probability

In this worksheet, students practise finding the probability of multiple conditional probabilities using the AND/OR statements.

'Apply Advanced Conditional Probability' worksheet

Key stage:  KS 4

Curriculum topic:  

Curriculum subtopic:  

Difficulty level:  

Worksheet Overview

QUESTION 1 of 10

There are multiple ways of finding probabilities of combined events. Sample space, two way tables and tree diagrams are all useful in their own way but what happens when you have 3 events, 4 events, n events?

 

When this happens, we have to use the AND/OR statements.

 

AND means we multiply

OR means we add

 

The trick with this is to rewrite the question with the words AND and OR.

 

Remember that these are conditional probabilities so the probabilities will change.

 

Example : A bag contains 7 black and 3 white balls. Three balls are removed without replacement. Find the probability that...

1) I get three black balls.

This can only happen one way...

black AND a black AND a black.

If we write this using fractions we get

7
10
x
6
9
x
5
8
=
210
630

This fraction will of course, cancel down to 1/3

 

2) I get 2 black and 1 white balls.

There are actually three different ways this can happen

black AND a black AND a white.

OR

black AND a white AND a black.

OR

A white AND a black AND a black.

The trick here is to know that each one of these probabilities will actually be the same (try it, you'll see it's true) so we only have to find one.

Lets find black AND a black AND a white.

7
10
x
6
9
x
3
8
=
126
630

We can now say that the probability of getting two blacks and a white is...

126
630
+
126
630
+
126
630
=
378
630

This will cancel down nicely to 3/5

 

3) The probability of getting at least one white.

For this, you could work out the probability of each one that satisfies this and add them all together (There's 7 ways this can happen)

or

You can use the fact that P(at least 1) = 1 - P(none)

So P(at least 1 white) = 1 - P(no whites)

P(at least 1 white) = 1 - P(3 blacks)

We worked out earlier that P(3 blacks) = 1/3

So we can say that the probability of getting at least one white = 1 - 1/3 = 2/3

At least probabilities can be found by...

6 red balls and 4 blue balls are in a bag.

I remove three balls without replacement.

Find the probability that I get...

 

3 red balls

 

Give your answer as a simplified fraction in the form a/b

6 red balls and 4 blue balls are in a bag.

I remove three balls without replacement.

Find the probability that I get...

 

Exactly one red ball

 

Give your answer as a simplified fraction in the form a/b

1/10

3/10

6 red balls and 4 blue balls are in a bag.

I remove three balls without replacement.

Find the probability that I get...

 

At least 1 blue ball

 

Using the at least formula, find the values os A and B for

1 - A = B

 

Give your answers as simplified fractions of the form a/b

 Value
A
B

In the word M A T H E M A T I C S, i pick three letters at random

Find the probability I get...

 

3 constonants

 

Give your answer as a simplified fraction in the form a/b

In the word M A T H E M A T I C S, i pick three letters at random

Find the probability I get...

 

Exactly two constanants

 

Give your answer as a simplified fraction in the form a/b

28/55

28/165

In the word M A T H E M A T I C S, i pick three letters at random

Find the probability I get...

 

At least one vowel

 

Using the at least formula, find the values os A and B for

1 - A = B

 

Give your answers as simplified fractions of the form a/b

 Value
A
B

In a deck of cards (without jokers), I pull out 3 random cards. Find the probability I get

 

3 hearts

 

Give your answer as a simplified fraction in the form a/b

In a deck of cards (without jokers), I pull out 3 random cards. Find the probability I get

Exactly two hearts

 

Give your answer as a simplified fraction in the form a/b

39/850

117/850

In a deck of cards (without jokers), I pull out 3 random cards. Find the probability I ge

 

At least one that isn't a heart

 

Using the at least formula, find the values os A and B for

1 - A = B

 

Give your answers as simplified fractions of the form a/b

  • Question 1

At least probabilities can be found by...

CORRECT ANSWER
EDDIE SAYS
This is an amazingly useful tool. It's really common with advanced probabilities that there will be a question at the start that asks you to find the probability of something happening (i.e. P 3 black balls) and another question that links to this at the end where it asks for a probability of at least 1 thing happening that is linked to this first question.
  • Question 2

6 red balls and 4 blue balls are in a bag.

I remove three balls without replacement.

Find the probability that I get...

 

3 red balls

 

Give your answer as a simplified fraction in the form a/b

CORRECT ANSWER
1/6
EDDIE SAYS
The only way this can happen is RED and RED and RED. All you need to do is find this probability which will be given by 6/10 x 5/9 x 4/8 and then cancel
  • Question 3

6 red balls and 4 blue balls are in a bag.

I remove three balls without replacement.

Find the probability that I get...

 

Exactly one red ball

 

Give your answer as a simplified fraction in the form a/b

CORRECT ANSWER
3/10
EDDIE SAYS
There are three ways this can happen RED and BLUE and BLUE or BLUE and RED and BLUE or BLUE and BLUE and RED We need to find the probability of one of these (remember they'll all come out the same) RED and BLUE and BLUE 6/10 x 4/9 x 3/8 = 1/10 1/10 + 1/10 + 1/10 = 3/10
  • Question 4

6 red balls and 4 blue balls are in a bag.

I remove three balls without replacement.

Find the probability that I get...

 

At least 1 blue ball

 

Using the at least formula, find the values os A and B for

1 - A = B

 

Give your answers as simplified fractions of the form a/b

CORRECT ANSWER
 Value
A
B
EDDIE SAYS
If we are using the at least formula. Remember that the probability of at least one is 1 - The probability of none In this case A is the probability I get no blue balls which means i am looking for the probability I find 3 reds. We found earlier that this was 1/6. This leaves B as 1 - 1/6
  • Question 5

In the word M A T H E M A T I C S, i pick three letters at random

Find the probability I get...

 

3 constonants

 

Give your answer as a simplified fraction in the form a/b

CORRECT ANSWER
7/33
EDDIE SAYS
The only way this can happen is CONST and CONST and CONAT All you need to do is find this probability which will be given by 7/11 x 6/10 x 5/9 and then cancel
  • Question 6

In the word M A T H E M A T I C S, i pick three letters at random

Find the probability I get...

 

Exactly two constanants

 

Give your answer as a simplified fraction in the form a/b

CORRECT ANSWER
28/165
EDDIE SAYS
There are three ways this can happen CONST and CONST and VOWEL or CONST and VOWEL and CONST or VOWEL and CONST and CONST We need to find the probability of one of these (remember they'll all come out the same) CONST and CONST and VOWEL 7/11 x 6/10 x 4/6 = 28/165 28/165 +28/165 + 28/165 = 28/55
  • Question 7

In the word M A T H E M A T I C S, i pick three letters at random

Find the probability I get...

 

At least one vowel

 

Using the at least formula, find the values os A and B for

1 - A = B

 

Give your answers as simplified fractions of the form a/b

CORRECT ANSWER
 Value
A
B
EDDIE SAYS
If we are using the at least formula. Remember that the probability of at least one is 1 - The probability of none In this case A is the probability I get no vowels which means i am looking for the probability I find 3 constonants. We found earlier that this was 7/33 This leaves B as 1 - 7/33
  • Question 8

In a deck of cards (without jokers), I pull out 3 random cards. Find the probability I get

 

3 hearts

 

Give your answer as a simplified fraction in the form a/b

CORRECT ANSWER
11/850
EDDIE SAYS
The only way this can happen is HEART and HEART and HEART All you need to do is find this probability which will be given by 13/52 x 12/51 x 11/50 and then cancel
  • Question 9

In a deck of cards (without jokers), I pull out 3 random cards. Find the probability I get

Exactly two hearts

 

Give your answer as a simplified fraction in the form a/b

CORRECT ANSWER
117/850
EDDIE SAYS
There are three ways this can happen HEART and HEART and OTHER or HEART and OTHER and HEART or OTHER and HEART and HEART We need to find the probability of one of these (remember they'll all come out the same) HEART and HEART and OTHER 13/52 x 12/51 x 39/50 = 39/850 39/850 + 39/850+ 39/850 = 117/850
  • Question 10

In a deck of cards (without jokers), I pull out 3 random cards. Find the probability I ge

 

At least one that isn't a heart

 

Using the at least formula, find the values os A and B for

1 - A = B

 

Give your answers as simplified fractions of the form a/b

CORRECT ANSWER
839/850
EDDIE SAYS
There's only one way this can't happen, which is if I get three hearts. We worked out earlier that this was 11/850 The probability of getting at least one that isn't a heart is 1 - P(all hearts) 1-11/850
---- OR ----

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