alleryStart learning →If a fair coin is tossed twice and $X$ = number of heads, then $P[X = 1]$ equals:
- $1/4$
- $3/4$
- $2/3$
- $1/2$
Show answer and solution
Correct answer: D. $1/2$
$X = 1$ corresponds to events {HT, TH}, so $P[X=1] = 2/4 = 1/2$.
MP DTE Probability Distribution — practice questions
Explore 29 free Probability Distribution questions in MP DTE Mathematics. Open each question or exercise to read its content, reference answers and available explanation.
Practise Probability Distribution with Quill →Probability Distribution questions and exercises with answers
Showing questions 1–29 of 29.
If a fair coin is tossed twice and $X$ = number of heads, then $P[X = 1]$ equals:Read the answer and worked solution →If two cards are drawn with replacement from a deck of 52, $P[X = 2]$ where $X$ = number of aces equals:Read the answer and worked solution →A fair die is thrown. Let $X$ = number of factors of the number on the upper face. $P[X = 2]$ equals:Read the answer and worked solution →Complete the definition of conditional probabilityRead the answer and worked solution →True or false: conditional probabilityRead the answer and worked solution →Independent or mutually exclusive?Read the answer and worked solution →Match each rule to its nameRead the answer and worked solution →IndependenceRead the answer and worked solution →Order the steps for a Bayes' theorem questionRead the answer and worked solution →Complete the conditions for a partitionRead the answer and worked solution →True or false: independence versus exclusivityRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Tap every statement that follows from independenceRead the answer and worked solution →The multiplication ruleRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Complete the theorem of total probabilityRead the answer and worked solution →Which theorem answers which question?Read the answer and worked solution →True or false: total probability and BayesRead the answer and worked solution →Match each historical figure to their contributionRead the answer and worked solution →The Historical NoteRead the answer and worked solution →Flip and pair: notation with meaningRead the answer and worked solution →Order the reasoning: from a partition to a posteriorRead the answer and worked solution →Tap every quantity that is a probabilityRead the answer and worked solution →Spot the odd one outRead the answer and worked solution →Assertion and reason: the multiplication ruleRead the answer and worked solution →True or false: the Historical NoteRead the answer and worked solution →Sort each statement by which idea it describesRead the answer and worked solution →
Worked practice questions
If two cards are drawn with replacement from a deck of 52, $P[X = 2]$ where $X$ = number of aces equals:
- $24/169$
- $1/52$
- $1/169$
- $144/169$
Show answer and solution
Correct answer: C. $1/169$
$P[\text{ace}] = 4/52 = 1/13$.
With replacement, draws are independent: $P[X=2] = (1/13)^2 = 1/169$.
A fair die is thrown. Let $X$ = number of factors of the number on the upper face. $P[X = 2]$ equals:
- $1$
- $1/6$
- $2/6$
- $3/6$
Show answer and solution
Correct answer: D. $3/6$
**Number of factors:** 1→1, 2→2 (1,2), 3→2 (1,3), 4→3 (1,2,4), 5→2 (1,5), 6→4 (1,2,3,6).
**So $X=2$ for {2,3,5} — 3 outcomes. $P[X=2] = 3/6 = 1/2$.**
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