A fair die is rolled six times. The probability of rolling "five or six" atleast five times is

Question

A fair die is rolled six times. The probability of rolling "five or six" atleast five times is (A) (12/729) (B) (12/(729)2) (C) (13/729) (D) (1/729)

Tardigrade Admin · Accepted Answer

P ( S )= P = P (5 or 6)=(1/3) P ( F )= q =(2/3) ; n =6 P text (atleast 5 text successes) = P (5)+ P (6)= 6 C 5((1/3))5((2/3))+ 6 C 6((1/3))6 =((1/3))6[12+1]=(13/729)

Q. A fair die is rolled six times. The probability of rolling "five or six" atleast five times is

$\frac{12}{729}$

$\frac{12}{( 729 ) ^{2}}$

$\frac{13}{729}$

$\frac{1}{729}$

$P (S) = P = P (5$ or 6 $) = \frac{1}{3}$
$P (F) = q = \frac{2}{3}; n = 6$
$P (atleast 5 successes) = P (5) + P (6) =^{6} C_{5} (\frac{1}{3})^{5} (\frac{2}{3}) +^{6} C_{6} (\frac{1}{3})^{6}$
$= (\frac{1}{3})^{6} [12 + 1] = \frac{13}{729}$

Q. A fair die is rolled six times. The probability of rolling "five or six" atleast five times is

72912​

(729)212​

72913​

7291​

P(S)=P=P(5 or 6)=31​ P(F)=q=32​;n=6 P (atleast 5 successes) =P(5)+P(6)=6C5​(31​)5(32​)+6C6​(31​)6 =(31​)6[12+1]=72913​

$\frac{12}{729}$

$\frac{12}{( 729 ) ^{2}}$

$\frac{13}{729}$

$\frac{1}{729}$

$P (S) = P = P (5$ or 6 $) = \frac{1}{3}$
$P (F) = q = \frac{2}{3}; n = 6$
$P (atleast 5 successes) = P (5) + P (6) =^{6} C_{5} (\frac{1}{3})^{5} (\frac{2}{3}) +^{6} C_{6} (\frac{1}{3})^{6}$
$= (\frac{1}{3})^{6} [12 + 1] = \frac{13}{729}$