What is the probability that the sum of the two numbers appearing on the top of the dice is less than or equal to 12?

Solution : When two different dice are thrown, the total number of outcomes is 36 and all the out-comes are equally likely. (i) The outcomes favourable to the event 'sum of two number is 9' are (6, 3), (5, 4), (4, 5) and (3, 6) these are 4 in number. `therefore" P (sum of 9) "=(4)/(36)=(1)/(9)`. (ii) The outcomes favourable to the event 'sum of two numbers is 10' are (6, 4), (5, 5) and ( 4, 6). These are 3 in number. `therefore" P (sum of 10) "=(3)/(36)=(1)/(12).` (iii) The sum of at least 10 mean that sum is 10, 11 or 12. Therefore, the outcomes favourable to the event 'sum of at least 10' are (6, 4), (5, 5), (4, 6), (6, 5), (5, 6) and ( 6, 6). These are 6 in number. `therefore" P (sum of at least 10) "=(6)/(36)=(1)/(6).` (iv) As the sum of two number appearing on the top of two dice can never be 13, there is no outcomes favourable to the event 'sum of two numbers is 13'. `therefore" P (sum of 13) "=(0)/(36)=0.` (v) As the sum of two numbers appearing on the top of two dice is always less than or equal to 12, all the 36 outcomes are favourable to the event 'sum is less than or equal to 12'. `therefore" P (sum is less than or equal to 12) "=(36)/(36)=1.` (vi) The outcomes favourable to the event 'a multiple of 2 on one die and a multiple of 3 on the other die' are (2, 3), (2, 6), (4, 3), (4, 6), (6, 3) (6, 6), (3, 2), (3, 4), (3, 6), (6, 2) and (6, 4). `therefore` The number of outcomes favourable to the given event = 11. `therefore" Required probability "=(11)/(36).`

Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:

less than or equal to 12

The number of possible outcomes = `6xx6=36`

All the outcomes are favourable to the event E=`"sum of two number"<=12:`

Hence, `p(E)=(n(E))/(n(S))=36/36=1`

Concept: Type of Event - Complementry

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