Given that, no of trials =n Probability of success (p)=
1
4
∴ Probability of no success =1−
1
4
=
3
4
As we know, probability of at least one success =1 - probability of no success ∴ According to the question, 1 - (Probability of no success in n trials )≥
9
10
⇒1−P(x=0)≥
9
10
⇒1−nC0(
1
4
)0(
3
4
)n≥
9
10
⇒1−(
3
4
)n≥
9
10
⇒(
3
4
)n≤
1
10
⇒(
4
3
)n≥10 [Taking log on both sides] ⇒n(log104−log103)≥log1010 ⇒n(log104−log103)≥1 ⇒n≥