Probability
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New answer posted
a month agoContributor-Level 10
n (s) = 6² = 36
E = { (1, 1), (1, 2), (1, 3), (1, 5), (1, 7), (2, 1), (2, 2), (2, 3), (2, 5), (3, 1), (3, 2), (3, 3), (3, 5), (5, 1), (5, 2), (5, 3), (7, 1)}
∴ n (E) = 17
Required prob. = n (E) / n (S) = 17 / 36
New answer posted
a month agoContributor-Level 10
11.00
Let probability of hitting the target = p ⇒ p=1/2
Let n be the minimum number of bombs
According to given condition
1 - (?C?P?(1-P)? + ?C?P¹(1-P)?¹) ≥ 99/100
⇒ 2? ≥ (n+1)100
n=10 ⇒ 2¹? ≥ 1100 Reject
n=11 ⇒ 2¹¹ ≥ 1200 Select
New answer posted
2 months agoContributor-Level 10
gen term =
For constant term
36 – 3r – rk = 0
for r = 1, 2, 4
12Cr212-r>28
Possible values of k = 3, 1
New answer posted
2 months agoContributor-Level 10
P (En) = n/36 for n = 1, 2, 3, …., 8
If one of the number from {1, 2, ….8} is left then total 29 by 3 ways
Similarly by leaving terms more 2 or 3 we get 16 more combinations
Total number of different set a possible is 16 + 3
= 19
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