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a) A sum 8

b) A doublet

c) A sum greater than 8

Answer

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Hint: Use the basic definition of probability of an event i.e. probability of any event is measured as the ratio of favourable cases to the whole number of cases. Find favourable cases for each problem given and use the above definition.

Here, it is given that two unbiased dice are rolled once and we need to find probability in all three cases provided.

As we know the basic definition of probability is the extent to which an event is likely to occur, measured by the ratio of the favourable cases to the whole number of cases possible.

Hence, probability of any event is given by;

$P=\dfrac{\text{Favourable cases to that event}}{\text{Total cases (Sample space)}}..........\left( 1 \right)$

We use the term ‘sample space’ as well for ‘total cases’.

Now, first we need to find sample space for rolling two unbiased dice. We have numbers on one dice as $\left\{ 1,2,3,4,5,6 \right\}$.

Here, we have two dices, so there will be pairs of numbers$\left\{ 1,2,3,4,5,6 \right\}$as given below;

Now, we have a table of sample space for rolling two unbiased dices.

Here, the total number of cases in sample space is 36 (6 x 6).

Now, let us calculate favourable cases for all three cases given in problem;

a) A sum 8

Let us calculate/ observe the total number of cases where the sum of numbers on first dice and second dice is 8.

So, favourable cases for this condition from the table given, we get;

$\left\{ \left( 2,6 \right),\left( 3,5 \right),\left( 4,4 \right),\left( 5,3 \right),\left( 6,2 \right) \right\}$

Hence, we get ‘5’ cases where the sum of numbers on both dice is 8.

Hence probability is given as;

$P=\dfrac{\text{Favourable cases}}{\text{Total cases (Sample space)}}$

We have,

Total cases = 36

Favourable cases = 5

$P=\dfrac{5}{36}$

Hence, probability of getting a sum 8 when two unbiased dice are rolled is $\dfrac{5}{36}$

b) A doublet

First we need to understand the term ‘doublet’ which is used when we have the same numbers on both dices.

Hence, favourable cases for this case would be that how many times, we get same numbers on dices;

Cases from table is given as;

$\left\{ \left( 1,1 \right),\left( 2,2 \right),\left( 3,3 \right),\left( 4,4 \right),\left( 5,5 \right),\left( 6,6 \right) \right\}$

Hence, number of favourable cases = 6

Therefore, Probability of getting doublet is given as;

$\begin{align}

& P=\dfrac{6}{36}=\dfrac{1}{6} \\

& P=\dfrac{1}{6} \\

\end{align}$

c) Sum greater than 8

Here, number of favourable cases from the table can be observed in a way such as;

Sum greater than 8 = number of cases for sum ‘9’ + number of cases for sum ‘10’ + number of cases for sum ‘11’ + number of cases for sum ‘12’.

Hence, favourable cases from the table of sample space, we get;

$\left\{ \left( 3,6 \right),\left( 4,5 \right),\left( 5,4 \right),\left( 6,3 \right),\left( 5,6 \right),\left( 6,5 \right),\left( 4,6 \right),\left( 5,5 \right),\left( 6,4 \right),\left( 6,6 \right) \right\}$

Hence, probability for sum greater than ‘8’ is given by;

$\begin{align}

& P=\dfrac{10}{36}=\dfrac{5}{18} \\

& P=\dfrac{5}{18} \\

\end{align}$

Hence, answers for the given problem are given as $\dfrac{5}{36},\dfrac{1}{6},\dfrac{5}{18}$.

Note: One can go wrong for writing the favourable cases for problem ‘3’ i.e. sum greater than 8. One can include cases for sum equal to ‘8’ as well which is wrong as the question is saying, we need to include sum greater than 8 only. Hence, we need to take care of these kinds of problems where ‘greater than’ or ‘lower than’ is involved.

One needs to prepare the whole table for solving these kinds of problems. We have provided a whole sample to make the solution easier and one can understand it easily.

One can go wrong when he/she misses any elements for favourable cases. Hence, always check twice or thrice for a favourable number of cases in probability.

Here, it is given that two unbiased dice are rolled once and we need to find probability in all three cases provided.

As we know the basic definition of probability is the extent to which an event is likely to occur, measured by the ratio of the favourable cases to the whole number of cases possible.

Hence, probability of any event is given by;

$P=\dfrac{\text{Favourable cases to that event}}{\text{Total cases (Sample space)}}..........\left( 1 \right)$

We use the term ‘sample space’ as well for ‘total cases’.

Now, first we need to find sample space for rolling two unbiased dice. We have numbers on one dice as $\left\{ 1,2,3,4,5,6 \right\}$.

Here, we have two dices, so there will be pairs of numbers$\left\{ 1,2,3,4,5,6 \right\}$as given below;

1 | 2 | 3 | 4 | 5 | 6 | |

1 | (1, 1) | (1, 2) | (1, 3) | (1, 4) | (1, 5) | (1, 6) |

2 | (2, 1) | (2, 2) | (2, 3) | (2, 4) | (2, 5) | (2, 6) |

3 | (3, 1) | (3, 2) | (3, 3) | (3, 4) | (4, 5) | (3, 6) |

4 | (4, 1) | (4, 2) | (4, 3) | (4, 4) | (4, 5) | (4, 6) |

5 | (5, 1) | (5, 2) | (5, 3) | (5, 4) | (5, 5) | (5, 6) |

6 | (6, 1) | (6, 2) | (6, 3) | (6, 4) | (6, 5) | (6, 6) |

Now, we have a table of sample space for rolling two unbiased dices.

Here, the total number of cases in sample space is 36 (6 x 6).

Now, let us calculate favourable cases for all three cases given in problem;

a) A sum 8

Let us calculate/ observe the total number of cases where the sum of numbers on first dice and second dice is 8.

So, favourable cases for this condition from the table given, we get;

$\left\{ \left( 2,6 \right),\left( 3,5 \right),\left( 4,4 \right),\left( 5,3 \right),\left( 6,2 \right) \right\}$

Hence, we get ‘5’ cases where the sum of numbers on both dice is 8.

Hence probability is given as;

$P=\dfrac{\text{Favourable cases}}{\text{Total cases (Sample space)}}$

We have,

Total cases = 36

Favourable cases = 5

$P=\dfrac{5}{36}$

Hence, probability of getting a sum 8 when two unbiased dice are rolled is $\dfrac{5}{36}$

b) A doublet

First we need to understand the term ‘doublet’ which is used when we have the same numbers on both dices.

Hence, favourable cases for this case would be that how many times, we get same numbers on dices;

Cases from table is given as;

$\left\{ \left( 1,1 \right),\left( 2,2 \right),\left( 3,3 \right),\left( 4,4 \right),\left( 5,5 \right),\left( 6,6 \right) \right\}$

Hence, number of favourable cases = 6

Therefore, Probability of getting doublet is given as;

$\begin{align}

& P=\dfrac{6}{36}=\dfrac{1}{6} \\

& P=\dfrac{1}{6} \\

\end{align}$

c) Sum greater than 8

Here, number of favourable cases from the table can be observed in a way such as;

Sum greater than 8 = number of cases for sum ‘9’ + number of cases for sum ‘10’ + number of cases for sum ‘11’ + number of cases for sum ‘12’.

Hence, favourable cases from the table of sample space, we get;

$\left\{ \left( 3,6 \right),\left( 4,5 \right),\left( 5,4 \right),\left( 6,3 \right),\left( 5,6 \right),\left( 6,5 \right),\left( 4,6 \right),\left( 5,5 \right),\left( 6,4 \right),\left( 6,6 \right) \right\}$

Hence, probability for sum greater than ‘8’ is given by;

$\begin{align}

& P=\dfrac{10}{36}=\dfrac{5}{18} \\

& P=\dfrac{5}{18} \\

\end{align}$

Hence, answers for the given problem are given as $\dfrac{5}{36},\dfrac{1}{6},\dfrac{5}{18}$.

Note: One can go wrong for writing the favourable cases for problem ‘3’ i.e. sum greater than 8. One can include cases for sum equal to ‘8’ as well which is wrong as the question is saying, we need to include sum greater than 8 only. Hence, we need to take care of these kinds of problems where ‘greater than’ or ‘lower than’ is involved.

One needs to prepare the whole table for solving these kinds of problems. We have provided a whole sample to make the solution easier and one can understand it easily.

One can go wrong when he/she misses any elements for favourable cases. Hence, always check twice or thrice for a favourable number of cases in probability.