The probability of getting a total of $9$ at least twice in $6$ tosses of a pair of dice is

^{6} C_{2} {(\frac{8}{9})}^{4} {(\frac{1}{9})}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{(\frac{8}{9})}^{6} +^{6} C_{1} {(\frac{8}{9})}^{5} {(\frac{1}{9})}^{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 - {(\frac{8}{9})}^{6} -^{6} C_{1} {(\frac{8}{9})}^{5} {(\frac{1}{9})}^{1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 - {(\frac{8}{9})}^{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $1 - {(\frac{8}{9})}^{6} -^{6} C_{1} {(\frac{8}{9})}^{5} {(\frac{1}{9})}^{1}$
For getting a total of $9$ in a pair of dice $= (3, 6), (6, 3), (4, 5), (5, 4)$
Probability of sum of $9$ in a pair of dice = $\frac{4}{36} = \frac{1}{9}$
$n = 9$
The probability of getting a total of $9$ at least twice $= 1 -$ [ the probability of not getting a total of $9 +$ The probability of getting a total of $9$ at once ]
= $1 -^{6} C_{0} {(\frac{1}{9})}^{0} {(1 - \frac{1}{9})}^{6 - 0} -^{6} C_{1} {(\frac{1}{9})}^{1} {(1 - \frac{1}{9})}^{6 - 1}$
$=$ $1 - {(\frac{8}{9})}^{6} -^{6} C_{1} {(\frac{1}{9})}^{1} {(\frac{8}{9})}^{5}$

Suggest Corrections