CUET UG 2025 — Mathematics Algebra

When tossing 3 coins simultaneously, let $x$x denote the number of heads obtained.

The condition $0 < x \leq 3$0<x≤3 means $x$x can be 1, 2, or 3 (at least one head).

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When tossing 3 coins, each coin shows either H (Head) or T (Tail).

Total possible outcomes $= 2 \times 2 \times 2 = 8$=2×2×2=8

All possible outcomes:

Outcome	Number of Heads (x)
HHH	3
HHT	2
HTH	2
HTT	1
THH	2
THT	1
TTH	1
TTT	0

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Outcomes where $0 < x \leq 3$0<x≤3 (i.e., $x = 1, 2,$x=1,2, or $3$3):

$x = 1$x=1: HTT, THT, TTH → 3 outcomes

$x = 2$x=2: HHT, HTH, THH → 3 outcomes

$x = 3$x=3: HHH → 1 outcome

Favorable outcomes $= 3 + 3 + 1 = 7$=3+3+1=7

Alternatively, only TTT has 0 heads, so favorable outcomes $= 8 - 1 = 7$=8−1=7

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$$P(0 < x \leq 3) = \dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$$P(0<x≤3)=Total outcomesFavorable outcomes​

$$P(0 < x \leq 3) = \dfrac{7}{8}$$P(0<x≤3)=87​

Therefore, the probability is $\dfrac{7}{8}$87​.