When a die is thrown twice,what is the probability that the sum of the numbers is $6$,given that at least one of the numbers is $4$?

  • A
    $1/6$
  • B
    $2/5$
  • C
    $3/5$
  • D
    $1/2$

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Similar Questions

$A$ and $B$ are two events such that $P(A) = 0.8$,$P(B) = 0.6$,and $P(A \cap B) = 0.5$. Then the value of $P(A/B)$ is:

Given $P(A)=0.5, P(B)=0.4, P(A \cap B)=0.3$,then $P(A^{\prime} / B^{\prime})$ is equal to

If $E_1$ and $E_2$ are two events of a random experiment such that $P(E_1) = \frac{1}{8}$,$P(E_1 \mid E_2) = \frac{1}{3}$,and $P(E_2 \mid E_1) = \frac{1}{4}$,then match the items of List-$I$ with the items of List-$II$.
List-$I$List-$II$
$A. P(E_1 \cup E_2)$$I. \frac{3}{29}$
$B. P(E_2)$$II. \frac{26}{29}$
$C. P(E_1 \mid \bar{E}_2)$$III. \frac{3}{16}$
$D. P(\bar{E}_1 \mid \bar{E}_2)$$IV. \frac{3}{32}$

$A$ and $B$ are independent events of a random experiment if and only if

Suppose that $E_1$ and $E_2$ are two events of a random experiment such that $P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$ and $P(E_1 / E_2) = \frac{1}{4}$. Observe the lists given below. The correct matching of List-$I$ with List-$II$ is:
List-$I$List-$II$
$(A)$ $P(E_2)$$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$$(iv)$ $1/2$
$(v)$ $3/8$
$(vi)$ $3/4$

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