In this chapter, we will solve questions related to throwing of two dice. We know that in a dice there are six numbers ( 1 to 6 ). When we throw two dice simultaneously, there are total of 6 x 6 = 36 possible outcomes. The sample space for experiment is given as;S = { (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) } The outcome of getting same number on both the dice is called doublets. (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) I hope you understood the basic concept. Let’s solve some problems; Questions on two dice probabilityQuestion (01) SolutionTotal possible outcome = 36We want to get (1, 1) or (1, 5) as final solution. Number of favorable outcome = 2 Probability = Number of favorable outcome / total possible outcome Probability = 2 / 36 = 1/18 Hence, 1/18 is the required probability Question 02 SolutionWhen two dice are thrown, the doublets are; (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) Number of favorable outcome = 6 Total possible outcomes = 36 Probability = 6 / 36 = 1 / 6 Hence, 1/6 is the required probability. Question 03In an experiment of throwing two dice simultaneously, find the probability for;(a) getting sum of 4 (b) getting sum of 9 Solution Probability = 3 / 36 = 1/12 (b) Let B be the event of getting sum 9 B = { (3, 6), (4, 5), (5, 4), (6, 3) } Number of favorable outcome = 4 Probability = 4 / 36 = 1/9 Hence, 1/9 is the required probability. Question 04 Solution A = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) } Number of favorable outcome = 18 Total possible outcome = 36 Probability = 18 / 36 = 1/2 Hence, 1/2 is the probability of getting even number on first dice. Question 05 (a) getting sum divisible by 7(b) getting sum at least 11 (c) getting even multiple on first and odd multiple on second dice Solution Probability = 6 / 36 = 1/6 (b) getting sum at least 11 Let B be the event of getting sum atleast 11. Number of favorable outcome = 3Total possible outcome = 36 Probability = 3/36 = 1/12 (c) getting even multiple on first and odd multiple on second dice C = {(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5),} Number of favorable outcome = 9Total possible outcome = 36 Probability = 9/36 = 1/4 Hence, 1/4 is the required probability. Hint: In this question, we are given that two dice are thrown simultaneously and we have to find various probabilities of numbers shown on both dice. For this, we will first make sample space and then use that to find favorable outcomes for finding each probability. Total outcomes will be given as the number of elements in sample space. Probability of any event is given as $\text{Probability}=\dfrac{\text{Number of favorable outcomes}}{\text{Total outcomes}}$. Complete step by step answer: Here, we are given that two dice are thrown simultaneously. As we know, a dice has 6 possibilities, therefore for two dice, the number of possibilities will be $6\times 6=36$. Let us draw sample space for the given event.(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)Hence, the total number of outcomes are 36.With the help of this sample space we will find required elements for every part.(i) Here we have to find the probability of getting the sum as 8. Therefore, let us analyze the sample space and count the numbers whose sum is 8. As we can see, following are required cases:(2,6), (3,5), (4,4), (5,3), (6,2)Hence, the number of favorable outcomes is 5. So,$\text{Probability}=\dfrac{5}{36}$.Hence, the probability of getting the sum as 8 is $\dfrac{5}{36}$. (ii) Let us analyze the sample space and count numbers whose one number is multiple of 2 i.e. 2, 4, 6 and another number is multiple of 3 i.e. 3, 6. As we can see, following are required cases:(2,3), (2,6), (4,3), (4,6), (6,3), (6,6), (3,2), (3,4), (3,6), (6,2), (6,4).Hence, the number of favorable outcomes is 11. So,$\text{Probability}=\dfrac{11}{36}$.Hence, probability of getting a multiple of 2 on one dice and multiple of 3 on other dice is $\dfrac{11}{36}$. (iii) Now, let us count numbers whose sum is at least 10, therefore, we have to count numbers whose sum is 10, 11 or 12. As we can see, following are the required cases:(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)Hence, the number of favorable outcomes are 6. So,$\text{Probability}=\dfrac{6}{36}=\dfrac{1}{6}$.Hence, the probability of getting a sum at least 10 is $\dfrac{1}{6}$. Note: Students should carefully count all the possibilities while calculating probability. In (ii) part, students should note that multiple of 2 or 3 can be on any of the two dices. For example, (2,3) and (3,2) both are favorable cases. In (iii) part, students should note that sum should be at least 10, so, they have to consider sum as 10 or higher. Try to avoid mistakes while making sample space. Two different dice are thrown at the same time. Find the probability of getting : (iii) a sum of at least 11 Total number of outcomes when two dice are rolled = 36 A Sum at least 11 favorable outcomes are: (5,6), (6,5), (6,6) Number of favorable outcomes = 3 P(Sum is at least 11) = `"Number of favourable outcomes "/"Total number of outcomes "` = `3/36 = 1/12` Concept: Simple Problems on Single Events Is there an error in this question or solution? |
Two dice are thrown simultaneously. find the probability of getting sum of at least 11
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