• # question_answer Two dice are thrown simultaneously. Match the probability of events in column-l to the column-ll. Column-I Column-II (P) Sum as prime number (i)   $\frac{11}{36}$ (Q) Multiple of 2 on one dice and  multiple of 3 on                      other dice                      (ii)  $\frac{1}{12}$ (R) Total of at least 10     (iii)  $\frac{5}{12}$ (S) Doublet of even numbers          (iv)  $\frac{1}{6}$ A)  P $\to$ (iv), Q $\to$ (i), R $\to$ (iii), S $\to$ (ii)B)  P $\to$ (iv), Q $\to$ (iii), R $\to$ (i). S $\to$ (ii)C)  P $\to$ (iii), Q $\to$ (ii), R $\to$ (iv), S $\to$ (i)D)  P $\to$ (iii), Q $\to$ (i), R $\to$ (iv), S $\to$ (ii)

Total number of outcomes when two dice are thrown$~=6\times 6=36$ (P) Possible ways for sum of numbers on dice is prime are (1, 1). (1,2), (1,4), (1,6), (2, 1), (2, 3), (2, 5), (3,2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 5), (6, 1) Favourable outcomes =15 $\therefore$ Required probability $=\frac{15}{36}=\frac{5}{12}$ (Q) Possible ways of multiple of 2 on one dice and multiple of 3 on other dice are (2, 3), (2, 6), (3, 2), (6, 2), (4, 3), (4, 6), (3, 4), (6, 4), (6, 3), (6, 6), (3, 6) Favourable outcomes =11 $\therefore$Required probability $=\frac{11}{36}$ (R) Possible ways of total of atleast 10 are (6, 6), (6, 5), (6, 4), (5, 6), (5, 5), (4, 6) Favourable outcomes = 6 $\therefore$   Required probability $=\frac{3}{36}=\frac{1}{12}$ (S) Possible ways for doublet of even numbers are (2, 2), (4, 4), (6, 6) Favourable outcomes = 3 $\therefore$  Required probability $=\frac{3}{36}=\frac{1}{12}$
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