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Probability

Question
CBSEENMA12033610

Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Solution

Here straight S space equals space open curly brackets table row cell left parenthesis 1 comma space 1 right parenthesis end cell cell left parenthesis 1 comma space 2 right parenthesis end cell cell left parenthesis 1 comma space 3 right parenthesis end cell cell left parenthesis 1 comma space 4 right parenthesis end cell cell left parenthesis 1 comma space 5 right parenthesis end cell cell left parenthesis 1 comma space 6 right parenthesis end cell row cell left parenthesis 2 comma space 1 right parenthesis end cell cell left parenthesis 2 comma space 2 right parenthesis end cell cell left parenthesis 2 comma space 3 right parenthesis end cell cell left parenthesis 2 comma space 4 right parenthesis end cell cell left parenthesis 2 comma space 5 right parenthesis end cell cell left parenthesis 2 comma space 6 right parenthesis end cell row cell left parenthesis 3 comma space 1 right parenthesis end cell cell left parenthesis 3 comma space 2 right parenthesis end cell cell left parenthesis 3 comma space 3 right parenthesis end cell cell left parenthesis 3 comma space 4 right parenthesis end cell cell left parenthesis 3 comma space 5 right parenthesis end cell cell left parenthesis 3 comma space 6 right parenthesis end cell row cell left parenthesis 4 comma space 1 right parenthesis end cell cell left parenthesis 4 comma space 2 right parenthesis end cell cell left parenthesis 4 comma space 3 right parenthesis end cell cell left parenthesis 4 comma space 4 right parenthesis end cell cell left parenthesis 4 comma space 5 right parenthesis end cell cell left parenthesis 4 comma space 6 right parenthesis end cell row cell left parenthesis 5 comma space 1 right parenthesis end cell cell left parenthesis 5 comma space 2 right parenthesis end cell cell left parenthesis 5 comma space 3 right parenthesis end cell cell left parenthesis 5 comma space 4 right parenthesis end cell cell left parenthesis 5 comma space 5 right parenthesis end cell cell left parenthesis 5 comma space 6 right parenthesis end cell row cell left parenthesis 6 comma space 1 right parenthesis end cell cell left parenthesis 6 comma space 2 right parenthesis end cell cell left parenthesis 6 comma space 3 right parenthesis end cell cell left parenthesis 6 comma space 4 right parenthesis end cell cell left parenthesis 6 comma space 5 right parenthesis end cell cell left parenthesis 6 comma space 6 right parenthesis end cell end table close curly brackets

Number of elements (outcomes) of the above sample space is 6 × 6 = 36.
Let E : the sum of numbers on the dice is 4
∴ E = {(1, 3), (2, 2), (3, 1)}
and F : numbers appearing on the two dice are different
∴ F contains all the points of S except (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
∴ F contains 30 elements.
Also E ∩ F = {(1, 3), (3, 1)}
therefore space space space space space space space straight P left parenthesis straight F right parenthesis space equals space 30 over 36 comma space space space straight P left parenthesis straight E intersection straight F right parenthesis space equals space 2 over 36
Required probability = straight P left parenthesis straight E space left enclose space straight F end enclose right parenthesis space equals space fraction numerator straight P left parenthesis straight E intersection straight F right parenthesis over denominator straight P left parenthesis straight F right parenthesis end fraction space equals fraction numerator begin display style 2 over 36 end style over denominator begin display style 30 over 36 end style end fraction space equals 2 over 36 cross times 36 over 30 space equals space 1 over 15

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