If p ^x = q, then?

Total number of outcomes = ¹⁰c₄ = 210
Favourable number of outcomes = ³c₂ x ²c₂ = 3 x 1 = 3
? Required probability = 3/210 = 1/70

You can have
W, W, B or W, B, W or B, W, W

Reqd. probability
= 3/4. 2/4. 3/4 + 3/4. 2/4. 1/4 + 1/4. 2/4. 1/4
= 9/32 + 3/32 + 1/32 = 13/32

Total balls in the box = 5
Second red ball can be drawn in two ways
Case I. First ball is white and second ball is red.
Its probability = 3/5.2/4 = 6/20 = 3/10
Case II: First ball is red and second ball is red
Its probability = 2/5.1/4 = 2/20 = 1/10
Hence, reqd. probability
= 3/10 + 1/10 = 4/10 = 2/5

Probability that trousers are not black = 2/3
Probability that shirts are not black = 3/4
Required robability = 2/3 x 3/4 = 1/2

n(S) = 2¹⁰⁰
n(E) = No. of favourable ways = ¹⁰⁰C₁ + ¹⁰⁰C₃ + ... ¹⁰⁰ C₉₉ = 2^100-1 = 2⁹⁹
[? ⁿC₁ + ⁿC₃ + ⁿC₅ + ........................... = 2^n-1 ]
? P(E)= n(E)/n(S) = 2⁹⁹/2¹⁰⁰ = 1/2

Note : The given case can be generalised as "If a unbiased coin is tossed 'n' times, then the chance that the head will present itself an odd number of times is 1/2. "

n(S) = ¹²C₃ = (12 x 11 x 10) / (3 x 2) = 2 x 11 x 10 = 220

No. of selecting of 3 oranges out of the total 12 orange = ⁴C₃ = 4
? n(E) = No. of desired selection of oranges = 220 - 4 = 216

? P(E) = n(E) / n(S) = 216 / 220 = 54/55

The total number of cause is 2¹⁰⁰.
The number of favourable cases are ¹⁰⁰C₁ + ¹⁰⁰C₃ + ..... + ¹⁰⁰C₉₉
= 2^{100 - 1} = 2⁹⁹
? Reqd probability = 2⁹⁹/ 2¹⁰⁰ = 1/2

P(M) = m, P (p) = p, P(c) = c
? The probability of at least one success
= P (M ? P ? C)
= m + p + c -mp - mc - pc + mcp = 3/4 ...(1)

The probability of at least two successes = mcp + mcp + mcp + mcp
= mc(1 - p) + mp (1 - c ) + (1 - m )cp + mcp
= mc + mp + cp - 2mcp = 1/2

The probability of exactly two success
= mcp + mcp + mcp
= mc(1 - p) + mp (1 - c ) cp(1 - m )
= mc + mp + cp - 3 mcp = 2/5

(2) & (3) gives,

? mcp = 1/2 - 2/5 = 1/10
? mc + mp + cp = 2/10 + 1/2 = 1/5 + 1/2 = 7/10


From (1),

m + p + c - 7/10 + 1/10 = 3/4
? m + p + c = 3/4 + 7/10 - 1/10 = 27/20
Thus, pmc = 1/10 is a true relation.

If the last digit in the product is to 2, 4, 6, 8 the last digit in all the n number should not be 0 and 5 and the last digit of all number should not be selected exclusively from the set of number {1, 3, 7, 9}
? Favourable number of cases
= 8ⁿ - 4ⁿ
But generally the last digit can be one of 0, 1, 2, 3, .... 9.
Hence, the total number of ways = 10ⁿ
Hence, the required probability
= 8ⁿ - 4ⁿ / 10ⁿ
= 4ⁿ - 2ⁿ / 5ⁿ

Let A = Event that A speaks the truth
and B = Event that B speaks the truth
Then, P(A) = 60/100 = 3/5 and P(A) = 80/100 = 4/5
? P(A) = (1 - 3/5) = 2/5 and
P(B) = (1 - 4/5) = 1/5

P(A and B contradict each other)
= P [(A speaks the truth and B tells a lie) or (A tells a lie and B speak the truth)]
=P[(A and B) or (A and B)]
=P(A and B) + P (A and B)
= P(A) x P(B) + P(A) x P(B)
= (3/5 x 1/5) + ( 2/5 x 4/5)
= (3/25 + 8/25 ) = 11/25
= (11/25 x 100)% = 44%