In a simultaneous throw of two coins, the probability of getting at least one head is:
View answer & explanation Ans: 3/4
Correct Answer: 3/4
Of the 4 outcomes HH, HT, TH, TT, 3 contain at least one head, so the probability is 3/4.
FPSC · Other · N/A
Solved probability MCQs — 19 questions on coins, dice, tickets, lottery draws and playing cards: at-least/at-most heads, dice totals and doublets, card draws and combined events, with verified answers and explanations. Free online practice for FPSC, PPSC, NTS and all one-paper exams. Covers probability.
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Questions
19
Time
19
Minutes
Subjects
1
Topics
3
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About this paper
This is the Arithmetic: Probability solved MCQ bank from Advanced FPSC & Other Federal Model Papers (Edition 32): 19 published questions (book pp. 773-775, of 20 printed) on coin tosses, dice totals and doublets, ticket and lottery draws, and single and paired card draws. Every answer was re-solved from scratch against the printed Answer Key, and every question carries a worked explanation.
Paper Controls
Subject 1
Topic Section
9 solved MCQs shown directly below
Correct Answer: 3/4
Of the 4 outcomes HH, HT, TH, TT, 3 contain at least one head, so the probability is 3/4.
Correct Answer: 1/2
Of the 8 outcomes, 4 have at least 2 heads (HHH, HHT, HTH, THH), so the probability is 4/8 = 1/2.
Correct Answer: 7/8
Only HHH has more than two heads, so 7 of the 8 outcomes qualify, giving 7/8.
Correct Answer: 1/3
The numbers greater than 4 are 5 and 6, so 2 of the 6 outcomes qualify, giving 2/6 = 1/3.
Correct Answer: 1/6
Six of the 36 pairs total 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1), so the probability is 6/36 = 1/6.
Correct Answer: 1/9
Four of the 36 pairs total 9 (3+6, 4+5, 5+4, 6+3), so the probability is 4/36 = 1/9. The second option is printed with a duplicate label (c); read by position as (b) 1/8.
Correct Answer: 1/6
Six of the 36 pairs are doublets (1,1 through 6,6), so the probability is 6/36 = 1/6.
Correct Answer: 5/36
Three pairs total 10 and two total 11, so 5 of the 36 outcomes qualify, giving 5/36.
Correct Answer: 3/4
The product is odd only when both dice are odd, which happens with probability 1/2 x 1/2 = 1/4, so the product is even with probability 1 - 1/4 = 3/4.
Topic Section
3 solved MCQs shown directly below
Correct Answer: 3/10
The multiples of 3 from 1 to 20 are 3, 6, 9, 12, 15 and 18, so 6 of 20 tickets qualify, giving 6/20 = 3/10.
Correct Answer: 9/20
Multiples of 3 (6) plus multiples of 5 (4) minus the overlap 15 (1) gives 9 of 20 tickets, so the probability is 9/20.
Correct Answer: 2/7
There are 10 prizes among 10 + 25 = 35 tickets, so the probability is 10/35 = 2/7.
Topic Section
7 solved MCQs shown directly below
Correct Answer: 1/26
Two named cards out of 52 qualify, so the probability is 2/52 = 1/26.
Correct Answer: 7/13
Red cards (26) plus kings (4) minus the two red kings gives 28 of 52, so the probability is 28/52 = 7/13.
Correct Answer: 4/13
Tens (4) plus spades (13) minus the ten of spades (1) gives 16 of 52, so the probability is 16/52 = 4/13.
Correct Answer: 4/13
Diamonds (13) plus kings (4) minus the king of diamonds (1) gives 16 of 52, so the probability is 16/52 = 4/13.
Correct Answer: 1/221
Choosing 2 kings from 4 over choosing 2 cards from 52 gives C(4,2)/C(52,2) = 6/1326 = 1/221.
Correct Answer: 13/102
There are 13 x 13 = 169 spade-heart pairs out of C(52,2) = 1326 pairs, so the probability is 169/1326 = 13/102.
Correct Answer: 55/221
Both red gives C(26,2) = 325 pairs and both kings gives C(4,2) = 6 pairs, with 1 pair (the two red kings) counted twice, so (325 + 6 - 1)/1326 = 330/1326 = 55/221.
Source & verification
Built by GK360 from verified bank questions — every answer rechecked before publishing.
Questions
19 questions, published of 20 printed in the book. One item was dropped, never inferred: the face-card question, whose standard probability 3/13 is not among the printed options and whose keyed answer 4/13 is wrong.
Coin tosses (at least one head, at least two heads, at most two heads), dice totals, sums and doublets, ticket and lottery draws, and card draws — single cards, named cards, and two-card combinations.
Yes. Every published answer was re-solved from scratch and checked against the book's printed Answer Key on p. 775; all 19 published answers agree with the key, and the one item where the key is wrong was dropped rather than published with a wrong answer.
Probability questions of this form appear in FPSC, PPSC, NTS, OTS and other one-paper exams, as well as CSS MPT-style quantitative sections, so the bank is tagged for all of them.
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