Total no . of cases = 6P2 = 6 x 5 = 30
Non-favourable cases are (4, 5) , (5, 4), (4, 6), (6, 4), (5, 6), (6, 5)
? Probability that event will not happen = 6 /(6 x 5) = 1/5
? Reqd. probability = 1 - 1/5 = 4/5
Probability of multiple of 2 = 3/6 = 1/2
Probability of multiple of 3 = 2/6 = 1/3
Since there are two dice.
? The required probability = 2 x 1/2 x 1/3 = 1/3
n(S) = 10C7 = 10C3
n(E) = 8C3
? P(E) = n(E)/n(S) = 7/15
? Probability of machine failing during a day p = 0.95
? q = Probability of its working during a day = 1 - p = 1 - 0.95 = 0.05
Required probability = q4 = (0.05)4
= 0.00000625
? Probability that no man out of n men aged x years will die in a year = (1 - p)n
? Probability that out of n men at least one will die in a year = 1 - (1 - p)n .
When at least one, man dies, any one out of the n men may be the first to die
Corresponding to each arrangement of (n - m) other books, there is a unique arrangement of the m volumes of the science book in ascending order and m! arrangement of the m volumes in random order
&Reqd. Prob = p = 1/m!
Let S be the sample space and E be the event of drawing 3 balls out of which 2 are blue and 1 is back .
Then, n (S)= Number of ways of drawing 3 balls out of 9=9c3
= (9 x 8 x 7) / (3 x 2 x 1)
= 84 and
n(E) = Number of ways of drawing 2 balls out of 5 and 1 ball out of 4 .
= 5C2 4C11
= (5 x 4) / (2 x 1) + 4 = 14
? P(E) = n(E)/n(S) = 14/84 = 1/6
? Probability for odd = p
? Probability for even = 2p
? p + 2p = 1
? 3p = 1
? p = 1/3
? Probability for odd = 1/3, Probability for even = 2/3.
Sum of two nos. is even means either both are odd or both are even
? Reqd. probability = 1/3 x 1/3 + 2/3 x 2/3 = 1/9 + 4/9 = 5/9
[ ? die is thrown twice ]
? P(A) = 5/9
? P(A) = 1- 5/9 = 4/9
? P(B) = 5/11
? P(B) = 1 - 5/11 = 6/11
Probability that none of them will occur = P(A ? B) = 4/9 x 5/ 11 = 20/99
Hence, Reqd. probability = 1-20/99 = 79//99
= 0.798 which is near 0.8.
The favourable cases are (1, 3), (2, 4) , (3, 5), (4, 6) and (1, 4), (2, 5), (3, 6) and their reversed cases like (3, 1), (4, 2), (5, 3)...
Total number of favourable cases = 2 x 7
? Required Probability P(E) = 14/36 = 7/18
Total number of cases is 17.
? Number divisible by 3 are 3, are 3, 6, 9, 12, 15 (These are 5 in number )
Number divisible by 7 are 7, 14. (These are 2 in number )
There are two favourable number of cases
Total no. of favourable number = 5 + 2
Required probability = 7/17.
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