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Three dice are thrown simultaneously. The probability that 4 has appeared on two dice given that 5 has occurred on one dice isa.1/6 b.3/91 c.3/216 d.none of these

Sviles Pool , 6 Years ago
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Saurabh Koranglekar

Last Activity: 6 Years ago

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Muskan soni

Last Activity: 5 Years ago

We need to know the favourable cases and the new sample space
No. Of pairs has to counted
(5,1,1) (5,1,2)............(5,1,6)
(5,2,1) (5,2,2)............(5,2,6)
.
.
.
..........................(5,6,6)
6×6=36 cases
 
Now considering series of 1
(1,1,1) (1,1,2).............(1,1,5) (1,1,6)
                                      (1,2,5)
                                      (1,3,5)
                                      (1,4,5)
(1,5,1)(1,5,2)(1,5,3)(1,5,4)(1,5,5)(1,5,6)
                                      (1,6,5)
11 pair sets in each series
5 such series of (1,2,3,4,6)
5 already counted having 36 sets
Total cases =5×11 +36
Total cases = 91(new sample space)
Favourable cases = (4,4,5)(4,5,4)(5,4,4)
Prob = 3/91

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