First we place seven white balls at places market W.
W -- W --W --W --W -- W -- W
If we place three black balls at -- places. Then no two black balls will be placed adjacently.
Total No, of -- places = 8
? No. of favourable ways for black balls = 8C3 = 8 x 7 x 6 / 1 x 2 x 3 = 56
Total no. of equally likely cases = 10! / 7!3!
= (10 x 9 x 8) / (1 x 2 x 3) = 15 x 8 = 120
? Reqd. Probability = 56 / 126 = 7/15
There are 13 hearts and 3 more kings
? p (heart or a king ) = (13 + 3)/52 = 4/13
? P (neither a heart nor a king) = 1 - 4/13 = 9/13
Clearly, n (S) = 52, there are 4 kings and 4 queens
? P(E) = n (E) / n (S) = 8/52 = 2/13
Clearly n(S) = 52. There are 26 red cards (including 2 kings ) and there are 2 more kings.
Let (E) be the event of getting either a red card or a king .
Then, n(E) = 28
? P(E) = n(E) / n(S) = 28/52 = 7/13
Clearly n(S) = 52 and there are 16 face cards.
? P(E) = 16/52 = 4/13
An ordinary year has 365 days i,e. 52 weeks and 1 day. So the probability that this day is a Sunday is 1/7.
? Probability for 3 = (1, 2), (2, 1) = 2/36
? Probability for 5 = (1, 4), (2, 3), (3, 2), (4, 1)= 4/ 36
? Probability for 11 (5, 6), (6, 5) = 2/36
? Reqd . Probability = 2/36 + 4/36 + 2/36 = 8/36 = 2/9
P(M) = 1/4; P(W)= 1/3
? P (M) = 1 - 1/4 = 3/4
? P(W)= 1 - 1/3 = 2/3
Reqd. probability = P(M) P(W) = 3/4 x 2/3 = 1/2
? Probability in each trial (shooting) = 0.3
? Reqd. probability = (0.3)10
? p = 1/5
? q = 1 - 1/5 = 4/5
The probability that none will hit in 10 shots = (4/5)10
? Reqd. probability = 1 - (4/5)10
Corresponding to n tosses, the Probability of getting no head = (1/2)n and,
Therefore, the probability of getting at least one head = 1 - (1/2)n
Now, 1 - (1/2)n ? 99/100
? (1/2)n ? 1/100
? n ? 7
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