# Ex.15.2 Q2 Probability Solution - NCERT Maths Class 10

## Question

A die is numbered in such a way that its faces show the numbers \(1, 2, 2, 3, 3, 6\). It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

What is the probability that the total score is (i) even? (ii) \(6\)? (iii) at least \(6\)?

## Text Solution

**What is known?**

A die is numbered in such a way that its faces show the numbers \(1, 2, 2, 3, 3, 6\). It is thrown two times and the total score in two throws is noted

**What is unknown?**

What is the probability that the total score is

(i) even?

(ii) \(6\)?

(iii) at least \(6\)?

**Reasoning:**

To solve this question, first find out the total number of outcomes and all the possible outcomes. Now, to find the probability use the formula given below

Probability

\[=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }\]

**Step:**

\(+\) | \(1\) | \(2\) | \(2\) | \(3\) | \(3\) | \(6\) |

\(1\) | \(2\) | \(3\) | \(3\) | \(4\) | \(4\) | \(7\) |

\(2\) | \(3\) | \(4\) | \(4\) | \(5\) | \(5\) | \(8\) |

\(2\) | \(3\) | \(4\) | \(4\) | \(5\) | \(5\) | \(8\) |

\(3\) | \(4\) | \(5\) | \(5\) | \(6\) | \(6\) | \(9\) |

\(3\) | \(4\) | \(5\) | \(5\) | \(6\) | \(6\) | \(9\) |

\(6\) | \(7\) | \(8\) | \(8\) | \(9\) | \(9\) | \(12\) |

Total number of possible outcomes \(= 6 \times 6 =36\)

(i) No of possible outcomes when the sum is even \(= 18\)

Probability that the total score is even

\[\begin{align} & =\frac{\text{ No of possible outcomes }}{\text{ Total no of outcomes }} \\& =\frac{18}{36}\\&=\frac{1}{2} \\\end{align}\]

(ii) No of possible outcomes when the sum is \(6 = 4\)

Probability that of getting the sum \(6\)

\[\begin{align} & =\frac{\text{ No of possible outcomes }}{\text{ Total no of outcomes }} \\& =\frac{4}{36}\\&=\frac{1}{9}\end{align}\]

(iii) No of possible outcomes when the sum is at-least6(greater than \(5\)) \(= 15\)

Probability that of getting the sum at-least \(6 \)

\[\begin{align} & =\frac{\text{No of possible outcomes }}{\text{Total no of outcomes}} \\& =\frac{15}{36}\\&=\frac{5}{12}\end{align}\]