In this section we consider the probabilities of equally likely events. When you roll a fair dice, each of the numbers 1 to 6 is equally likely to be on the uppermost face of the dice.

For equally likely events:

p(a particular outcome) | = |

A card is taken at random from a full pack of 52 playing cards. What is the probability that it is:

Note: As each card is equally likely to be drawn from the pack there are 52 equally likely outcomes.

(a)

a red card,

There are 26 red cards in the pack, so:

p(red) | = | = |

(b)

a 'Queen',

There are 4 Queens in the pack, so:

p(Queen) | = | = |

(c)

a red 'Ace',

There are 2 red Aces in the pack, so:

p(red Ace) | = | = |

(d)

the 'Seven of Hearts',

There is only one 7 of Hearts in the pack, so:

p(7 of Hearts) | = |

(e)

an even number?

There are 20 cards that have even numbers in the pack, so:

p(even number) | = | = |

A packet of sweets contains 18 *red* sweets, 12 *green* sweets and 10 *yellow* sweets. A sweet is taken at random from the packet. What is the probability that the sweet is:

The total number of sweets in the packet is 40, so there are 40 equally likely outcomes when one is taken at random.

(a)

*red*,

There are 18 *red* sweets in the packet, so:

p(red) | = | = |

(b)

*not green*,

There are 28 sweets that are *not green* in the packet, so:

p(not green) | = | = |

(c)

*green* or *yellow* ?

There are 22 sweets that are *green* or *yellow* in the packet, so:

p(green or yellow) | = | = |

You roll a fair dice 120 times. How many times would you expect to obtain:

(a)

a 6,

p(6) | = |

Expected number of 6s | = | × 120 | = | 20 |

(b)

an *even* score,

p(even score) | = | = |

Expected number of even scores | = | × 120 | = | 60 |

(c)

a score of *less than* 5 ?

p(score less than 5) | = | = |

Expected number of less than 5 | = | × 120 | = | 80 |