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Miscellaneous Exercise · Q3

Q.A die has two faces each with number '1', three faces each with number '2' and one face with number '3'. If die is rolled once, determine

(i) P(2)
(ii) P(1 or 3)
(iii) P(not 3).
Rajasthan RbseTextbookSubjective· 2mImportance★★★★★
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✓ Free question

A non-standard die with 6 faces (two 1's, three 2's, one 3) gives probabilities by counting favorable outcomes: P(2)=12P(2) = \frac{1}{2}, P(1 or 3)=12P(1 \text{ or } 3) = \frac{1}{2}, and P(not 3)=56P(\text{not } 3) = \frac{5}{6}.

The heart of probability is counting what you want versus counting everything possible. When a die is rolled, each face has an equal chance of landing up. This die isn't standard—it has repeated numbers—but the six physical faces are still equally likely. So we count faces, not distinct numbers.

The sample space has 6 equally likely outcomes (the six faces), distributed as:

  • Two faces show '1'
  • Three faces show '2'
  • One face shows '3'

For any event, the probability is the ratio of favorable faces to total faces.


(i) Finding P(2)P(2)

  1. Identify favorable outcomes: We want the die to show a '2'. Three of the six faces have '2' written on them.

  2. Apply the probability formula:

P(2)=Number of faces showing 2Total number of faces=36=12P(2) = \frac{\text{Number of faces showing 2}}{\text{Total number of faces}} = \frac{3}{6} = \frac{1}{2}


(ii) Finding P(1 or 3)P(1 \text{ or } 3)

  1. Understand the "or" operation: The event "11 or 33" means the die shows either a '1' or a '3'. In set language, this is the union of two events: {1}∪{3}\{1\} \cup \{3\}.

  2. Count favorable faces:

    • Faces showing '1': 2 faces
    • Faces showing '3': 1 face
    • Total favorable: 2+1=32 + 1 = 3 faces
  3. Calculate the probability:

P(1 or 3)=36=12P(1 \text{ or } 3) = \frac{3}{6} = \frac{1}{2}

Tip

For mutually exclusive events (a face can't show both '1' and '3' simultaneously), P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B). Here: P(1)+P(3)=26+16=36P(1) + P(3) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6}.


(iii) Finding P(not 3)P(\text{not } 3)

  1. Interpret "not 3": This is the complement of the event "getting a 3". The die must show either '1' or '2'.

  2. Count favorable faces:

    • Faces showing '1': 2 faces
    • Faces showing '2': 3 faces
    • Total favorable: 2+3=52 + 3 = 5 faces
  3. Calculate directly:

P(not 3)=56P(\text{not } 3) = \frac{5}{6}

Alternatively, using the complement rule:

P(not 3)=1−P(3)=1−16=56P(\text{not } 3) = 1 - P(3) = 1 - \frac{1}{6} = \frac{5}{6}

Watch out

Don't confuse the number of distinct values with the number of faces. The die has 3 distinct numbers but 6 faces. Always count physical faces when computing probabilities.


✓Final answer

The probabilities are: (i) P(2)=12P(2) = \boxed{\frac{1}{2}}, (ii) P(1 or 3)=12P(1 \text{ or } 3) = \boxed{\frac{1}{2}}, and (iii) P(not 3)=56P(\text{not } 3) = \boxed{\frac{5}{6}}.

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