If an unbiased die, marked with -2,-1,0,1,2,3 on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is :

Solution

Maths sir · 3 steps

Sort each face as positive, negative or zero. Then you only need an even number of negatives and no zero.

1. Sort the faces. Three faces are positive (1, 2, 3), two are negative (−2, −1) and one is zero. On each throw, P(positive) = 1/2, P(negative) = 1/3 and P(0) = 1/6.
2. When is it positive. A single 0 makes the product zero, so every throw must be non-zero. The sign is then positive exactly when the number of negatives is even: 0, 2 or 4 of the five throws.

   P = Σ_(k=0,2,4) C(5, k)(1/3)ᵏ(1/2)⁵⁻ᵏ

   Negatives k	Probability	Over 2592
   0	(1/2)⁵ = 1/32	81/2592
   2	10 × 1/9 × 1/8 = 5/36	360/2592
   4	5 × 1/81 × 1/2 = 5/162	80/2592

3. So. P = (81 + 360 + 80)/2592 = 521/2592, which is option (b).

The idea to keep. For the sign of a product, ignore the sizes of the numbers. Treat each trial as positive, negative or zero, then use the binomial formula over an even count of negatives.

Answer from the NTA final answer key. Solution by PrepWiser.