If two dice are thrown simultaneously find the probability that the numbers appearing are equal

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Question 17 Two dice are thrown simultaneously 500 times. Each time the sum of two numbers appearing on their tops is noted and recorded as given in the following table begin array | c | c | Whine |text Sum & Frequency Whhine 2 & 14 lllhline 3 & 30 whline 4 & 42 Whline5 & 55 Whline6 & 72 lllhline7 & 75 ll hline8 & 70 llhline9 & 53 Whline10 & 46 ilvhline11 & 28 Whline12 & 15 Illhline end array If the dice are thrown once more, then what is the probability of getting a sumi 3 ?ii More than 10 ?iii Less than or equal to 5 ?iv Between 8 and 12 ?

Solution : When two different dice are thrown, the total number of outcomes is 36 and all the out-comes are equally likely. <br> (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. <br> `therefore" P (sum of 9) "=(4)/(36)=(1)/(9)`. <br> (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. <br> `therefore" P (sum of 10) "=(3)/(36)=(1)/(12).` <br> (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. <br> `therefore" P (sum of at least 10) "=(6)/(36)=(1)/(6).` <br> (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'. <br> `therefore" P (sum of 13) "=(0)/(36)=0.` <br> (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'. <br> `therefore" P (sum is less than or equal to 12) "=(36)/(36)=1.` <br> (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). <br> `therefore` The number of outcomes favourable to the given event = 11. <br> `therefore" Required probability "=(11)/(36).`

Solution:

Given, two dice are thrown simultaneously.

We have to find the probability that the sum of the numbers appearing on the dice is 1.

When 2 dice are thrown at the same time, the overall possible outcomes are

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

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

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

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

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

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

Total number of possible outcomes = 36

Number of favourable outcomes = 0

Number of possible outcomes = 36

Probability = number of favourable outcomes / number of possible outcomes

Probability of getting a sum of 1 = 0/36

Therefore, the probability of getting the sum of 1 on the dice is zero.

✦ Try This: Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is 0?

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 14

NCERT Exemplar Class 10 Maths Exercise 13.3 Problem 20(iii)

Summary:

Two dice are thrown simultaneously. The probability that the sum of the numbers appearing on the dice is 1 is zero

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