• Space complexity is a measure of how efficient your code is in terms of memory used.

• Since the product has to end in 1, 3, 7 or 9, the units place of all the 4 numbers we want to multiply can’t end in 0, 2, 4, 5, 6, 8. That means it can only end in 1, 3, 7, 9.
• This means that the probability of selecting one number that ends in 1, 3, 7 or 9 is 4/10 = 2/5.
• Therefore the probability of selecting four numbers that end in 1, 3, 7 or 9 is (2/5) x (2/5) x (2/5) x (2/5) = 16/625.
• Hence, the probability of selecting 4 numbers such that their product ends in 1, 3, 7 or 9 is 16/625.

Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) + (Prob. that brother is not selected) × (Prob. that sister is selected)

Hence, Required Probability = 1/5 x 2/3 + 4/5 x 1/3
= 2/15 + 4/15
= 6/15
= 2/5.

• n(S) = Number of methods for drawing 2 balls out of 13 = 13C2
  = (13 x 12) / 2
  = 78.
• n(E) = No. of methods for drawing 2 balls out of 5 = 5C2
  = (5 x 4) / 2
  = 10.
• Therefore, P(E) = n(E) / n(S)
  = 10/78
  = 5/39.

n(S) = 36
n(E) = {(5,6), (6,5)} = 2
p(E) = n(E) / n(S)
= 2/36
= 1/18.

Total number of cards n(S) = 52
Number of red cards n(E) = 26.
Hence, P(E) = n(E) / n(S)
= 26/52
= 1/2.

P(A) = 7/8, P(B) = 9/10
Required Probability = P(A) x P(B)
= (7/8) x (9/10)
= 63/80.

n(S) = 52.
Let E = Event of getting a queen of club or a king of heart.
Then, n(E) = 2.
P(E) = n(E) / n(S)
= 2 / 52
= 1 / 26

Probability of getting head in one trail = 1/2.
Therefore, Required Probability of getting heads in both the trails = 1/2 x 1/2 = 1/4.

Here, S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
P(E) = n(E)/n(S)
= 7/8.

log 125 / log 5 = x.
log 5³ / log 5 = x
x = 3 log5 / log5
= 3

log₂(1/64)
= log₂(1) – log₂(64)
= 0 – log₂(2⁶)
= -6log₂(2).
= −6

Here are some values to base ten represented as logarithms :

10^0.3010 = 2 ,
take log base 10 both sides ,
log₁₀10^0.3010 = log₁₀2
0.3010 = log₁₀2

Given :
log_0.125125 = log_125/1000125
= log_0.55 = 1 / log₅0.5
= 1 / -log₅2
= -log₁₀5 / log₁₀2
= -2.321 [ log5 = 0.6989 and log2 = 0.3010]

logX + logY = log (XY)
= log (a²/bc) + log (b²/ac) + log (c²/ab)
= log (a² b² c² / bcacab)
= log (a² b²c² / a²b²c²)
= log 1
= 0

Given:
= log75/16 – 2 log5/9 + log32/343
= log [(25 x 3)/(4 x 4)] – log (25/81) + log [(16 x 2)/(81 x 3)] = log(25 x 3) – log ( 4 x 4 ) – log(25) + log81 + log(16 x 2) – log (81 x 3)
= log 25 + log 3 – log 16 – log 25 + log 81 + log 16 + log 2 – log 81 – log 3
= log 2

Given:
= log [ { (9 / 8) / (27 / 32) } x 3 / 4) ] = log [ (9 / 8) x (3 / 4) x (32 / 27) ] = log 1
= 0

log(ab²) – log(ac) + log abc⁴ – 3log(bc)
= log ab² – logac + log abc⁴ – log b³c³
= log_a (ab²/ac) x (abc⁴/b³c³)
= log_aa = 1

log₂x = 10 ⇒ x = 2¹⁰.
log_xY= 100
⇒ y = x¹⁰⁰
⇒ y = (2¹⁰)¹⁰⁰ [put value of x] ⇒ y = 2¹⁰⁰⁰

Let required number of hours be x
More examiners, less hours(indirect proportion)
More days, less hours(indirect proportion)
More answer books, more hours(direct proportion)
Hence we can write as

• 9 * 12 * 2 * 5 = 4 * 30 * 1 * x

• x = 9 * 12 * 2 * 5 / 4 * 30 * 1

• = 9 * 3 * 2 * 5 / 30

= 9