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:

Question

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: (A) (440/2592) (B) (881/2592) (C) (27/288) (D) (521/2592)

Tardigrade Admin · Accepted Answer

Either all outcomes are positive or any two are negative.

Now, p = P (positive) = \frac{3}{6} = \frac{1}{2}

q = p (negative) = \frac{2}{6} = \frac{1}{3}

Required probability

=^{5} C_{5} (\frac{1}{2})^{5} +^{5} C_{2} (\frac{1}{3})^{2} (\frac{1}{2})^{3} +^{5} C_{4} (\frac{1}{3})^{4} (\frac{1}{2})^{1}

= \frac{521}{2592}

Q. 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:

$\frac{440}{2592}$

$\frac{881}{2592}$

$\frac{27}{288}$

$\frac{521}{2592}$

Q. 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:

2592440​

2592881​

28827​

2592521​

Either all outcomes are positive or any two are negative. Now, p=P( positive )=63​=21​ q=p( negative )=62​=31​ Required probability =5C5​(21​)5+5C2​(31​)2(21​)3+5C4​(31​)4(21​)1 =2592521​

Q. 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:

$\frac{440}{2592}$

$\frac{881}{2592}$

$\frac{27}{288}$

$\frac{521}{2592}$