A and B throw a die alternatively till one of them gets a number more than 4 and wins the game.
Then the probability of winning the game by B, if A starts first:

Which among the following is NOT a corner point of the LPP, maximize Z = 4x + y subject to the constraints $$x + y \le 50, 3x + y \le 90, x \ge 0, y \ge 0 ?$$​​

If the corner points of the LPP, $$Z = 13x − 15y$$​ subject to the constraints $$x + y \le 7, 2x − 3y + 6 \ge 0, x \ge 0, y \ge 0$$​​
are (0,0), (7,0), (3,4) and (0,2), then the value of Maximum Z + 3 Minimum Z is:

Three numbers chosen from 1 to 20. Probability that they are not consecutive is:

The image of point (−1, 3, 4) in the plane $$x − 2y = 0$$​ is:

For any vector $$\vec{a},$$​ the value of $$(\vec{a} \times \hat{i})^{2} + (\vec{a} \times \hat{j})^{2} + (\vec{a} \times \hat{k})^{2}$$​ is:

Which among the following is NOT a corner point of the LPP, maximize Z = 4x + y subject to the constraints $$x + y \le 50, 3x + y \le 90, x \ge 0, y \ge 0 ?$$​​

If the corner points of the LPP, $$Z = 13x − 15y$$​ subject to the constraints $$x + y \le 7, 2x − 3y + 6 \ge 0, x \ge 0, y \ge 0$$​​
are (0,0), (7,0), (3,4) and (0,2), then the value of Maximum Z + 3 Minimum Z is:

Three numbers chosen from 1 to 20. Probability that they are not consecutive is:

The image of point (−1, 3, 4) in the plane $$x − 2y = 0$$​ is:

For any vector $$\vec{a},$$​ the value of $$(\vec{a} \times \hat{i})^{2} + (\vec{a} \times \hat{j})^{2} + (\vec{a} \times \hat{k})^{2}$$​ is: