Solution When two dice are thrown , total possible outccomes=36 Favorable outcomes for 4 will not come up either time: { (1,1) (1,2) (1,3) (1,5) (1,6)} { (2,1) (2,2) (2,3) (2,5) (2,6)} { (3,1) (3,2) (3,3) (3,5) (3,6)} { (5,1) (5,2) (5,3) (5,5) (5,6)} { (6,1) (6,2) (6,3) (6,5) (6,6)} Number of favorable outcomes = 25 p (4 will not come up) = 25/36

Step-by-step explanation: Probability of getting 1 as uppermost face = 3/216 or 1/72. Probability of getting 2 as uppermost face = 3/216 or 1/72. Probability of getting 3 as uppermost face = 3/216 or 1/72. Probability of getting 4 as uppermost face = 3/216 or 1/72. Probability of getting 5 as uppermost face.

Find the probability of getting a larger number than the previous number each time. According to the question, we have . n = 6, r = 3. Total number of sample space when die is rolled 3 times = 6 3 [∵ Total number of sample space when die is rolled r times = n r] Number of favourable outcomes = C 3 6