Tag: some more terms in probability

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Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

The probability that an event does not happens in one trial is 0.8.The probability that the event happens atmost once in three trails is 

  1. $0.896$
  2. $0.791$
  3. $0.642$
  4. $0.592$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The probability of failure in one trial is q = 0.8, so success p = 0.2. For n=3 trials, the probability of at most one success is P(X=0) + P(X=1). P(X=0) = (0.8)^3 = 0.512. P(X=1) = 3 * (0.2)^1 * (0.8)^2 = 3 * 0.2 * 0.64 = 0.384. Summing these gives 0.512 + 0.384 = 0.896.

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

If for two events $A$ and $B, P(A\cap B)\ne P(A) \times P(B)$, then the two events $A$ and $B$ are

  1. Independent

  2. Dependent

  3. Not equally likely

  4. Not exhaustive
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For independent.events $P\left( A\cap B \right) =P\left( A \right) .P\left( B \right) $

So, $P\left( A\cap B \right) \neq P\left( A \right) .P\left( B \right) $ implies that A and B are independent.

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

A bag contains four tickets marked with $112, 121, 211, 222$, one ticket is drawn at random from the bag. Let $E _i(i=1, 2, 3)$ denote the event that $i^{th}$ digit on the ticket is $2$ then :

  1. $E _1$ and $E _2$ are independent
  2. $E _2$ and $E _3$ are independent
  3. $E _3$ and $E _1$ are independent
  4. $E _1, E _2, E _2$ are independent
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

$P(E _1) = P(E _2) = P(E _3) =\dfrac{1}{2}$


$P(E _i \cap E _j) = \dfrac{1}{4} = P(E _i)P(E _j)$

Hence, two events taken together are independent.

$P(E _1 \cap E _2 \cap E _3) = \dfrac{1}{4} \neq P(E _1)P(E _2)P(E _3)$

Therefore, the three events are not independent together.

Hence, options A, B and C are correct.

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

Two cards are drawn simultaneously from a well shuffled pack of $52$ cards. The expected number of aces is?

  1. $\dfrac{1}{221}$
  2. $\dfrac{3}{131}$
  3. $\dfrac{2}{113}$
  4. $\dfrac{1}{131}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{matrix} Two\, \, cards\, \, are\, \, drawn\, \, from\, \, a\, \, well\, \, shuffered\, \, pack\, \, of\, \, 52\, \, card. \ Then,\, \, number\, \, access\, \, is\, \,  \ \Rightarrow number\, \, of\, \, sample\, \, space=52/2 \ \Rightarrow There\, \, are\, \, four\, \, ace=\dfrac { { 4/2 } }{ { 52/2 } } =\frac { { 4\times 3 } }{ { 52\times 51 } }  \ =\dfrac { 1 }{ { 13\times 17 } } \, \, \, \, \, Ans. \  \end{matrix}$

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

If P(A) = P(B), then

  1. A and B are the same events

  2. A and B must be same events

  3. A and B may be different events

  4. A and B are mutually exclusive events.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given $P(A) = P(B)$
Then $A$ can be different event from $B$ because $A\;\xi\;B$ are Mutually exclusive.
i.e, $A\;\xi\;B$ may be different events.
Hence, the answer is $A$ and $B$ may be different events.