Two dice are thrown together 4 times. The probability that both dice will show same numbers twice is -

Question

Two dice are thrown together 4 times. The probability that both dice will show same numbers twice is - (A) (1/3) (B) (25/36) (C) (25/216) (D) None of these

Tardigrade Admin · Accepted Answer

The probability of showing same number by both dice

p = \frac{6}{36} = \frac{1}{6}

In binomial distribution here

n = 4, r = 2, p = \frac{1}{6}, q = \frac{5}{6}

∴

req. probability =

^{n} C_{r} q^{n - r} p^{r} =^{4} C_{2} (\frac{5}{6})^{2} (\frac{1}{6})^{2}

= 6 (\frac{25}{36}) (\frac{1}{36}) = \frac{25}{216}

Q. Two dice are thrown together $4$ times. The probability that both dice will show same numbers twice is -

$\frac{1}{3}$

$\frac{25}{36}$

$\frac{25}{216}$

None of these

Q. Two dice are thrown together 4 times. The probability that both dice will show same numbers twice is -

31​

3625​

21625​

None of these

The probability of showing same number by both dice p=366​=61​ In binomial distribution here n=4,r=2,p=61​,q=65​ ∴ req. probability = nCr​qn−rpr=4C2​(65​)2(61​)2 =6(3625​)(361​)=21625​

Q. Two dice are thrown together $4$ times. The probability that both dice will show same numbers twice is -

$\frac{1}{3}$

$\frac{25}{36}$

$\frac{25}{216}$