Face | 1 | 2 | 3 | 4 | 5 | 6 |
P | 0.01 | 0.32 | 0.21 | 0.15 | 0.05 | 0.147 |
A) \[\frac{16}{21}\]
B) \[\frac{1}{10}\]
C) \[\frac{5}{16}\]
D) \[\frac{5}{21}\]
Correct Answer: D
Solution :
[d] Let E: ?face 1 comes up? and F: ?face 1 or 2 comes up? \[\Rightarrow E\cap F=E\] \[(\therefore E\subset F)\] \[\therefore P(E)=0.10\] and \[P(F)=P(1)+P(2)=0.10+0.32=0.42\] Hence, required probability \[=P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{P(E)}{P(F)}=\frac{0.10}{0.42}=\frac{5}{21}\]You need to login to perform this action.
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