Probability Test
Probability
This is Probability Test-01 for CBSE class 12 Maths.. There are 15 questions in this test with each question having around four answer choices.
Questions & Answers
1
The conditional probability of an event E, given the occurrence of the event F is given by
- A\({\text{P}}({\text{E}}|{\text{F}}){\text{}} = \frac{{P\left( {E\mathop \cap \nolimits^ F} \right)}}{{P\left( E \right)}},\;P(F) \ne 0\)
- B$\text{P}(\text{E }\!\!~\!\!\text{ }|\text{ }\!\!~\!\!\text{ F})\text{ }\!\!~\!\!\text{ }=\frac{P\left( E\mathop{\cap }^{}F \right)}{P\left( F \right)},~P(F)\ne 0$Correct
- C\({\text{P}}({\text{E}}|{\text{F}}){\text{}} = \frac{{P\left( {E\mathop \cap \nolimits^ F} \right)}}{{P\left( F \right)}},\;P(F) < 0\)
- D\({\text{P}}({\text{E}}|{\text{F}}){\text{}} = \frac{{P\left( {E\mathop \cup \nolimits^ F} \right)}}{{P\left( F \right)}},\;P(F) \ne 0\)
2
The conditional probability of an event E, given the occurrence of the event F
- A${\text{0 < P (E|F) < 1}}$
- B0 ≤ P (E|F) < 1
- C0 ≤ P (E|F) ≤ 1Correct
- D0 < P (E|F) ≤ 1
3
The conditional probability of an event E’s complement E’, given the occurrence of the event F
- AP (E′|F) = 1 – P (E|F)Correct
- BP (E′|F) = – 1 + P (E|F)
- CP (E′|F) = P (E|F)
- DP (E′|F) = 1
4
If E, F and G are events then P ((E ∪ F)|G) =
- AP (E|G) + P (F|G) – P ((E ∩ F)|F)
- BP (E|G) + P (G|F) – P ((E ∩ F)|G)
- CP (G|E) + P (F|G) – P ((E ∩ F)|E)
- DP (E|G) + P (F|G) – P ((E ∩ F)|G)Correct
5
If E and F are events then P (E ∩ F) =
- AP (E ∩ F) P (F|E), P (E) ≠ 0
- BP (E) P (E|F), P (E) ≠ 0
- CP (E∪F) P (F|E), P (E) ≠ 0
- DP (E) P (F|E), P (E) ≠ 0Correct
6
Two coins are tossed once ,where E : tail appears on one coin , F : one coin shows head. Find P(E/F).
- A0.23
- B0.24
- C0.33
- D1Correct
7
Two coins are tossed once ,where E :no tail appears , F : no head appers. Find P(E/F).
- A0.24
- B0Correct
- C0.25
- D0.22
8
Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E)
- AP (E|F) =\(\frac{2}{3}\),P(F|E) = \(\frac{1}{4}\)
- BP (E|F) =\(\frac{2}{3}\),P(F|E) = \(\frac{1}{3}\)Correct
- CP (E|F) =\(\frac{2}{4}\),P(F|E) = \(\frac{1}{3}\)
- DP (E|F) =\(\frac{2}{5}\),P(F|E) = \(\frac{1}{4}\)
9
Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32
- AP(A|B) = \(\frac{{16}}{{29}}\)
- BP(A|B) = \(\frac{{15}}{{27}}\)
- CP(A|B) = \(\frac{{16}}{{33}}\)
- DP(A|B) = \(\frac{{16}}{{25}}\)Correct
10
If P (A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find P(A ∩ B)
- A0.32Correct
- B0.37
- C0.29
- D0.35
11
If P (A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find P(A|B)
- A0.68
- B0.66
- C0.62
- D0.64Correct
12
If P (A) = 0.8, P (B) = 0.5 and P(B|A) = 0.4, find P(A ∪ B)
- A0.25
- B1.00
- C0.95
- D0.98Correct
13
Evaluate P(A ∪ B), if 2P(A) = P(B) =\(\frac{5}{{13}}\) and P(A|B) =\(\frac{2}{5}\)
- A\(\frac{{11}}{{29}}\)
- B\(\frac{{11}}{{26}}\)Correct
- C\(\frac{{15}}{{26}}\)
- D\(\frac{{11}}{{27}}\)
14
If P(A) =\(\frac{6}{{11}}\) , P(B) =\(\frac{5}{{11}}\)and P(A ∪ B) = \(\frac{7}{{11}}.\)find P(A|B)
- A\(\frac{3}{7}\)
- B\(\frac{4}{5}\)Correct
- C\(\frac{4}{5}\)
- D\(\frac{3}{7}\)
15
If P(A) =\(\frac{6}{{11}}\) , P(B) =\(\frac{5}{{11}}\)and P(A ∪ B) = \(\frac{7}{{11}}.\)find P(B|A)
- A\(\frac{3}{5}\)
- B\(\frac{4}{5}\)
- C\(\frac{2}{3}\)Correct
- D\(\frac{1}{3}\)